All Solutions 1078

Instructor’s Manual

Essential Mathematics forEconomic Analysis

2nd edition

Knut SydsæterPeter Hammond

For further instructor material please visit:

www.pearsoned.co.uk/sydsaeter

ISBN 13: 978-0-273-68185-4/ISBN 10: 0-273-68185-0

© Knut Sydsæter and Peter Hammond 2006Lecturers adopting the main text are permitted to download the manual as required

Pearson Education LimitedEdinburgh GateHarlowEssex CM20 2JEEngland

and Associated Companies throughout the world

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First published 2006

© Knut Sydsæter and Peter Hammond 2006

The rights of Knut Sydsæter and Peter Hammond to be identified as authors of this workhave been asserted by them in accordance with the Copyright, Designs and Patents Act 1988.

ISBN 13: 978-0-273-68185-4/ISBN 10: 0-273-68185-0

All rights reserved. Permission is hereby given for the material in thispublication to be reproduced for OHP transparencies and student handouts,without express permission of the Publishers, for educational purposes only.In all other cases, no part of this publication may be reproduced, storedin a retrieval system, or transmitted in any form or by any means, electronic,mechanical, photocopying, recording or otherwise, without either the priorwritten permission of the publisher or a licence permitting restricted copyingin the United Kingdom issued by the Copyright Licensing Agency Ltd,90 Tottenham Court Road, London W1T 4LP. This book may not be lent,resold, hired out or otherwise disposed of by way of trade in any form ofbinding or cover other than that in which it is published, without theprior consent of the publishers.


Preface

This instructor’s and solutions manual accompanies Essential Mathematics for Economic Analysis (2nd edi-tion, FT Prentice Hall, 2006). Its main purpose is to provide solutions to the even-numbered problems.(Answers to the odd-numbered problems are given in the main text.) For many of the more interesting and/ordifficult problems, detailed solutions are provided—more detailed than instructors are likely to need for them-selves. Nevertheless, the extra detail may be useful if answers are supplied to the students, or as a guide tomarking answers. Sometimes only an outline of the solution is given. For some of the simpler problems, onlythe final answer is presented.

In addition, for each chapter we have a section with comments on the content. Sometimes we explainwhy certain topics are included and others are excluded. There are also occasional useful hints based on ourexperience of teaching the material. In some cases, we also comment on alternative approaches, sometimeswith mild criticism of other ways of dealing with the material that we believe to be less suitable.

Chapters 1 and 2 in the main text review elementary algebra. This manual includes a Test I, designed forthe students themselves to see if they need to review particular sections of Chapters 1 and 2.

Many students using our text will probably have some background in calculus. The accompanying Test IIis designed to give information to both the students and the instructors about what students actually knowabout single variable calculus, and about what needs to be studied more closely, perhaps in Chapters 6 to 9 ofthe text.

Oslo and Stanford, August 2005

Knut Sydsæter, Peter Hammond

Contact addresses:[email protected]@stanford.edu


Contents

1 Introductory Topics I: Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2 Introductory Topics II: Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

3 Introductory Topics III: Miscellaneous . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

4 Functions of One Variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

5 Properties of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

6 Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

7 Derivatives in Use . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

8 Single-Variable Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

9 Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

10 Interest Rates and Present Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

11 Functions of Many Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

12 Tools for Comparative Statics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

13 Multivariable Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

14 Constrained Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

15 Matrix and Vector Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67


16 Determinants and Inverse Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

17 Linear Programming . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

Test I (Elementary Algebra) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

Test II (Elementary Mathematics) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

Answers to Test I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

Answers to Test II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87


6 C H A P T E R 2 I N T R O D U C T O R Y T O P I C S I I : E Q U A T I O N S

Chapter 1 Introductory Topics I: Algebra

The main purpose of Chapter 1 is to help those students who need to revise elementary algebra. (Those whonever learned it will need more help than we can provide here.)

Section 1.1 briefly discusses the real number system and distinguishes between natural numbers, integers,rational numbers, and irrational numbers.

Section 1.2 concerns rules for manipulating powers, starting with an explanation of the notation, and therules that a0 = 1, a−n = 1/an (if a �= 0). Then fundamental properties of exponents are described. Next, itis shown how powers can be used to calculate compound interest. Finally, negative powers are applied to findhow much money should have been deposited in an account a given number of years ago in order to yield atarget amount today.

Section 1.3 concerns algebraic rules. Twelve of the most common rules are presented first, and illustratedin examples. One is even illustrated geometrically. (A surprisingly large number of students seem unawareof how algebraic rules can be illustrated in this way.) There follow three identities concerning squares. Thestudent is told what to do with a minus sign outside parentheses. The anatomy of algebraic expressions isdefined—terms in particular, and numerical coefficients.

Fractions may have frightened some students in the past. Section 1.4 may help to overcome the inducedamnesia. Notation is explained. Common factors are cancelled. Denominators are made rational. “Stupid”errors are described. Rules for signs are explained. Algebraic fractions are added, and the rules illustratednumerically. The use of LCDs (lowest common denominators) is illustrated. Parentheses are suggested wherethey help avoid errors. Finally come rules for multiplying and dividing fractions. All very elementary, butcrucially important, and sources of common errors among students.

Section 1.5 explains fractional powers and begins with the well-known case of square roots. Fractionalpowers are important in economics as will be shown in later chapters. (See Section 4.8.)

The next Section 1.6 discusses inequalities, which are often not much discussed in elementary algebracourses. Notation and some elementary properties start the section. It is shown how to “solve” linear inequal-ities. Among the properties presented is the often neglected (5)—if each side of an inequality is multipliedby a negative number, the inequality gets reversed. Then sign diagrams are introduced as a useful device forseeing when certain products or quotients are positive, and when they are negative. The section concludeswith a brief discussion of double inequalities.

Intervals on the real line, absolute values, and the usual measure of distance are discussed in Section 1.7.All the problems in Chapter 1 are supplied with answers in the text itself to make this chapter more

suitable for self study.

Chapter 2 Introductory Topics II: Equations

Mathematics for economic analysis often involves solving equations, which is the topic of this chapter.In Section 2.1 even the simplest linear equations receive the attention that some students may well need.

Nonlinear equations can lead to attempts to divide by zero, which must be carefully avoided. Often, thehardest part of solving a problem is in formulating an appropriate equation. This skill is hard to teach, exceptby numerous examples. Some practice can come from the problems.


C H A P T E R 3 I N T R O D U C T O R Y T O P I C S I I I : M I S C E L L A N E O U S 7

Section 2.2 discusses equations with parameters which arise in many economic applications, yet are oftenneglected in textbooks. The problems for this section are aimed at training students to solve such equations.

Then Section 2.3 covers quadratic equations (in a single variable). It describes the crucially importanttechnique of completing the square, as well as how to use the roots of a quadratic equation to factor theleft-hand side. Only after several examples is the technique described in general. Then the expressions forthe sum and product of the two roots (when they are real) are given.

Section 2.4 deals with the solution of two equations in two unknowns. First, the nature of the problem isexplained. Then two methods of solution are proposed. The first method involves using one equation to solvefor one variable in terms of the other. After this, the solution for both variables is easily derived by substitution,of course. The second method is to eliminate one of the variables by adding or subtracting multiples of oneequation from the other. An appropriate tableau for doing this is described. The section ends with a formulagiving the solution to a general pair of two linear equations in two unknowns. More general simultaneouslinear equations, together with techniques for solving them, receive extended coverage in Chapters 15 and 16.

Section 2.5 gives some examples of how to solve some types of nonlinear equation which frequentlyoccur in optimization problems.

All the problems in this chapter are supplied with answers in the text itself to make this material moresuitable for self study.

Chapter 3 Introductory Topics III: MiscellaneousThe chapter discusses some miscellaneous mathematical fundamentals. Section 3.1 explains summation nota-tion, a topic that often causes difficulties to the untutored. Simple examples illustrate the general notation. Itis important to understand that the index of summation is a “dummy variable”, and also to understand whatto do with indices that are not indices of summation.

Then Section 3.2 presents some rules for manipulating sums, including sums of sums and commonfactors. An important example in which most terms cancel is presented. So are formulas connected with the(arithmetic) mean and variance of a set of numbers. Some common errors are also noted. Then the formulafor the sum of an arithmetic progression is presented. The formula is generalized in Problem 3.2.5. We alsogive the formulas for the sums of the first n squares and first n cubes. Next comes the binomial formula.Pascal’s triangle is displayed (and attributed more appropriately).

Section 3.3 gives a brief coverage of double sums. There is an analogy with a standard spreadsheet. Themain point is that summing each row and then summing the row totals is equivalent to summing each columnand then summing the column totals. In other words, the order of summation does not matter.

Section 3.4 discusses some aspects of logic. Propositions are explained, as are implications. Also, neces-sary and sufficient conditions are clearly distinguished from each other. The use of logic to solve equationsand inequalities is illustrated.

Next, Section 3.5 discusses proofs, including both direct proofs (that P ⇒ Q) and indirect proofs (thatnot Q ⇒ not P ). Also mentioned is the distinction between deductive and inductive reasoning.

A brief exposition of set theory comes in Section 3.6. Sometimes sets can be defined by listing theirmembers. More generally, they can be defined by the common property or properties each of their membersmust satisfy. Standard notation for set membership, unions, intersections, differences, the empty set, andcomplements is introduced. So are Venn diagrams, which are then applied to show some easy results.

Finally, the last Section 3.7 of this chapter treats proof by (mathematical) induction. It is an importantidea to grasp.


8 C H A P T E R 3 I N T R O D U C T O R Y T O P I C S I I I : M I S C E L L A N E O U S

Answers to Even-Numbered Problems

3.12. (a) 2

√0 + 2

√1 + 2

√2 + 2

√3 + 2

√4 = 2(3 + √

2 + √3)

(b) (x + 0)2 + (x + 2)2 + (x + 4)2 + (x + 6)2 = 4(x2 + 6x + 14)

(c) a1ib2 + a2ib

3 + a3ib4 + · · · + anib

n+1 (d) f (x0)�x0 + f (x1)�x1 + f (x2)�x2 + · · · + f (xm)�xm

4.2 · 3 + 3 · 5 + 4 · 7

1 · 3 + 2 · 5 + 3 · 7· 100 = 6 + 15 + 28

3 + 10 + 21· 100 = 49

34· 100 ≈ 144.12

6. (a) The total number of people moving from region i. (b) The total number of people moving to region j .

3.22. (a + b)6 = a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6. (The coefficients are those in the

seventh row of Pascal’s triangle in the text.)

4. (a)

(5

3

)= 5 · 4 · 3

1 · 2 · 3= 5 · 4 · 3 · 2 · 1

1 · 2 · 3 · 2 · 1= 5!

3! 2!= 5!

2! 3!. In general,

(m

k

)= m(m − 1) · · · (m − k + 1)

k!

= m(m − 1) · · · (m − k + 1) · (m − k)!

k!(m − k)!= m!

(m − k)! k!.

(b)

(8

3

)= 8 · 7 · 6

1 · 2 · 3= 56. Also,

(8

8 − 3

)=

(8

5

)= 8 · 7 · 6 · 5 · 4

1 · 2 · 3 · 4 · 5= 56;

(8

3

)+(

8

3 + 1

)=

56 + 8 · 7 · 6 · 5

1 · 2 · 3 · 4= 126 and

(8 + 1

3 + 1

)=(

9

4

)= 9 · 8 · 7 · 6

1 · 2 · 3 · 4= 126.

(c)

(m

k

)= m!

(m − k)!k!=(

m

m − k

)and

(m

k

)+(

m

k + 1

)= m!

(m − k)!k!+ m!

(m − k − 1)!(k + 1)!

= m!(k + 1 + m − k)

(m − k)!(k + 1)!= (m + 1)!

(m − k)!(k + 1)!=(

m + 1

k + 1

)3.3

2. (a) The total number of units of good i.(b) The total number of units of all goods owned by person j .(c) The total number of units of goods owned by the group as a whole.

4. Because arj − a is independent of the summation index s, it is a common factor when we sum over s, so∑ms=1(arj − a)(asj − a) = (arj − a)

∑ms=1(asj − a) for each r . Next, summing over r gives

m∑r=1

m∑s=1

(arj − a)(asj − a) =[ m∑

r=1

(arj − a)

][ m∑s=1

(asj − a)

](∗∗)

Using the properties of sums and the definition of aj , we have

m∑r=1

(arj − a) =m∑

r=1

arj −m∑

r=1

a = maj − ma = m(aj − a)

Similarly, replacing r with s as the index of summation, one also has∑m

s=1(asj − a) = m(aj − a).Substituting these values into (∗∗) then confirms (∗).


C H A P T E R 3 I N T R O D U C T O R Y T O P I C S I I I : M I S C E L L A N E O U S 9

3.4

2. x = 2. (x = −1, 0, and 1 make the equation meaningless. Multiplying each term by the commondenominator x(x − 1)(x + 1) yields 2x(x2 − 3x + 2) = 0, or 2x(x − 1)(x − 2) = 0. Hence, x = 2 isthe only solution.)

4. (a) x ≥ 0 is necessary, but not sufficient. (b) x ≥ 50 is sufficient, but not necessary.(c) x ≥ 4 is necessary and sufficient.

6. (a) No solutions. (Squaring each side yields x−4 = x+5−18√

x + 5+81, which reduces to√

x + 5 = 5,with solution x = 20. But x = 20 does not satisfy the given equation.)(b) x = 20

8. (a) x+√x + 4 = 2 ⇒ √

x + 4 = 2−x ⇒ x+4 = 4−4x+x2 ⇒ x2−5x = 0(i)⇒ x−5 = 0

(ii)⇐ x = 5.Here implication (i) is incorrect (x2 − 5x = 0 ⇒ x − 5 = 0 or x = 0.) Implication (ii) is correct, butit breaks the chain of implications. (b) x = 0. (After correcting implication (i), we see that the givenequation implies x = 5 or x = 0. But only x = 0 is a solution; x = 5 solves x − √

x + 4 = 2.)

3.5

2. (a) Logically the two statements are equivalent. (b)Appending the second statement is still an expressivepoetic reinforcement.

3.6

2. F ∩B ∩C is the set of all female biology students in the university choir; M ∩F the female mathematicsstudents;

((M ∩B)\C

)\T the students who study both mathematics and biology but neither play tennisnor belong to the university choir.

4. (a) B ⊂ M (b) F ∩ B ∩ C �= ∅ (c) T ∩ B = ∅ (d) F \ (T ∪ C) ⊂ B

6. (b) and (c) are true, the others are wrong. (Counter example for (a), (d), and (f): A = {1, 2}, B = {1},C = {1, 3}. As for (e), note in particular that A ∪ B = A ∪ C = A whenever B and C are subsets of A,even if B �= C.)

8. (a) Consider Fig. M3.6.8, and let nk denote the number of people in the set marked Sk , for k = 1, 2, . . . , 8.Obviously n1 + n2 + · · · + n8 = 1000. The responses imply that: n1 + n3 + n4 + n7 = 420; n1 + n2 +n5 + n7 = 316; n2 + n3 + n6 + n7 = 160; n1 + n7 = 116; n3 + n7 = 100; n2 + n7 = 30; n7 = 16,and n8 = 334. From these equations we easily find n1 = 100, n2 = 14, n3 = 84, n4 = 220, n5 = 186,n6 = 46, n7 = 16. (i) n3 + n4 = 304 had read A but not B; (ii) n6 = 46; (iii) n8 = 334.(b) We find n(A\B) = n3 +n4 = 304, n(C \ (A∪B)) = n6 = 46, and n(�\ (A∪B ∪C)) = n8 = 334.The last equality is a special case of n(� \ D) = n(�) − n(D). (The number of persons who are in �,but not in D, is the number of persons in all of � minus the number of those who are in D.)


10 C H A P T E R 3 I N T R O D U C T O R Y T O P I C S I I I : M I S C E L L A N E O U S

S4

S1

S7

S3

S2

S5

S6S8

A B

C

Figure M3.6.8

3.72. We prove only (3.2.6); the proof of (3.2.5) is very similar, but slightly easier. For n = 1 the LHS and the

RHS of (3.2.6) are both equal to 1. As the induction hypothesis, suppose (3.2.6) is true for n = k, so that

k∑i=1

i3 = 13 + 23 + · · · + k3 = [ 12k(k + 1)]2

Thenk+1∑i=1

i3 =k∑

i=1

i3 + (k + 1)3 = [ 12k(k + 1)]2 + (k + 1)3 = (k + 1)2( 1

4k2 + k + 1)

But this last expression is equal to 14 (k + 1)2(k2 + 4k + 4) = [ 1

2 (k + 1)(k + 2)]2, which proves that(3.2.6) is true for n = k + 1. By induction, we have proved (3.2.6).

4. The claim is true for n = 1. As the induction hypothesis, suppose k3 + (k + 1)3 + (k + 2)3 is divisibleby 9. Note that (k + 1)3 + (k + 2)3 + (k + 3)3 = (k + 1)3 + (k + 2)3 + k3 + 9k2 + 27k + 27 =k3 + (k +1)3 + (k +2)3 +9(k2 +3k +3). This is divisible by 9 because the induction hypothesis impliesthat the sum of the first three terms is divisible by 9, whereas the last term is also obviously divisible by 9.

Review Problems for Chapter 3

2. (a) 12·4+22·5+32·6+42·7 = 4+20+54+112 = 190 (b) 1− 16 = 5

6 (c) 1−2+2−1+30+41+52+63 =1 + 1/2 + 1 + 4 + 25 + 216 = 495/2

4. (a)38∑i=4

i(i + 2) (b)n∑

i=1

1

xi(c)

16∑j=0

x2j

2j + 1(d)

81∑k=1

(−1)k−1 1

k

6. (a) ⇒ true, ⇐ false (b) ⇒ false (because x2 = 16 also has the solution x = −4), ⇐ true (c) ⇒ true,⇐ false (x can be 3) (d) ⇒ and ⇐ both true

8. A ∩ B = ∅; A ∪ B = {1, 2, 4, 6, 11, 12}; � \ B = {1, 3, 4, 5, 6, 7, 8, 9, 10}; CA = � \ A ={2, 3, 5, 7, 8, 9, 10, 11, 12}

10. For n = 1, the inequality is correct: 1 + x ≥ 1 + x. Suppose (1 + x)k ≥ 1 + kx for an arbitrary naturalnumber k. We have to prove that then (1 + x)k+1 ≥ 1 + (k + 1)x. But because x ≥ −1 and k > 0, wehave (1 + x)k+1 = (1 + x)k(1 + x) ≥ (1 + kx)(1 + x) = 1 + (k + 1)x + kx2 ≥ 1 + (k + 1)x. Thus, byinduction, Bernoulli’s inequality is true for all natural numbers n.


C H A P T E R 4 F U N C T I O N S O F O N E V A R I A B L E 11

Chapter 4 Functions of One Variable

Functions of one real variable are the main subject of Chapters 4 to 9. Chapter 4 begins with a gentleintroduction. Section 4.1 presents a few simple examples, mostly drawn from economics, before Section 4.2presents a definition, and introduces functional notation systematically. The domain and range of a functionare defined, as are increasing and decreasing functions.

In our experience, training students to master functional notation is very important, yet sometimes neg-lected in introductory texts. The students should be encouraged to do many of the numerous problems in thissection, and throughout the chapter.

Next, Section 4.3 begins with a review of how functions can be represented by graphs in a two-dimensionalcoordinate system. It displays the graphs of a number of important simple functions which students need tolearn as soon as possible.

The special case of linear functions is the subject of Section 4.4. The intercept is defined, and the formulafor the slope provided. Next come the “point–slope” formula for the straight line with a given slope througha specified point, and the “point–point” formula for the line through two specified points with different x-coordinates. The section concludes with the general equation for a straight line that includes the case when theline is vertical, along with the graphical method of solving linear equations, and the graphical representationof linear inequalities.

Section 4.5 considers a (dubious) linear model of population growth and the linear consumption functionas well as a simple model of market equilibrium of supply and demand.

Section 4.6 treats quadratic functions, and the possible shapes of their graphs. The solution of quadraticequations is reviewed (see also Section 2.3) and the method of completing the square is used to identify themaximum or the minimum point of a quadratic function. The Cauchy–Schwarz inequality is discussed as aharder problem.

In order to have students see mathematics applied as soon as possible to economic optimization problems,we consider a monopolist with a quadratic cost function who faces a linear (inverse) demand function. Theperfectly competitive firm is treated as a special case. It is shown that, in order that the profit maximum fora monopolist should coincide with that for a competitive firm, its output should be subsidized (rather thantaxed). (Alternatively, it should have its price regulated.)

Next, Section 4.7 covers more general polynomials, starting with cubic functions. An example showshow they can represent a richer family of cost curves. The solutions of polynomial equations are discussed.Techniques for finding integer roots are emphasized. Then division of polynomials is considered, especiallythe remainder theorem. The correspondence between the roots and linear factors of a polynomial is pointedout. Then a technique for dividing polynomials is presented. Finally, rational functions—both proper andimproper—are defined.

Power functions are the subject of Section 4.8. The need for fractional powers is illustrated, and rationalpowers are then defined. The graphs of power functions are illustrated.

Section 4.9 treats exponential functions. The natural exponential function is introduced already at thisstage, although the reason for its importance will only be clear once students have mastered some calculus.Chapter 4 concludes with a discussion of logarithmic functions, emphasizing the natural logarithmic functionwhich is now the most commonly used.


12 C H A P T E R 4 F U N C T I O N S O F O N E V A R I A B L E


4.22. F(0) = F(−3) = 10, F(a + h) − F(a) = 10 − 10 = 0

4. (a) f (−1/10) = −10/101, f (0) = 0, f (1/√

2) = √2/3, f (

√π) = √

π/(1 + π), f (2) = 2/5(b) f (−x) = −x/(1 + (−x)2) = −x/(1 + x2) = −f (x) and f (1/x) = (1/x)/[1 + (1/x)2] =(1/x) · x2/[1 + (1/x)2] · x2 = x/(1 + x2) = f (x)

6. (a) C(0) = 1000, C(100) = 41 000, and C(101) − C(100) = 501(b) C(x + 1) − C(x) = 2x + 301 = incremental cost of increasing production from x to x + 1.

8. (a) f (tx) = 100(tx)2 = 100t2x2 = t2f (x) (b) P(tx) = (tx)1/2 = t1/2x1/2 = t1/2P(x)

10. (a) No. f (2 + 1) = f (3) = 18, whereas f (2) + f (1) = 10. (b) Yes. f (2 + 1) = f (2) + f (1) = −9(c) No. f (2 + 1) = f (3) = √

3 ≈ 1.73, whereas f (2) + f (1) = √2 + 1 ≈ 2.41.

12. See Figs. M4.2.12a and M4.2.12b.

x2

x · 1 1 · 1

1 · x

x 1

x

1

x

1

x

1

Figure M4.2.12a The areais (x + 1)2 = x2 + 2x + 1

Figure M4.2.12b The areais x2 + 1

14. (a) Defined for x �= 2, i.e. Df = (−∞, 2) ∪ (2, ∞) (b) f (8) = 5

(c) f (x) = 3x + 6

x − 2= 3 ⇐⇒ 3x + 6 = 3(x − 2) ⇐⇒ 6 = −6, which is impossible.

4.32. (a) f (−5) = 0, f (−3) = −3, f (−2) = 0, f (0) = 2, f (3) = 4, f (4) = 0

(b) Df = [−5, 4], Rf = [−3, 4]

4.x −2 −1 0 1 2 3 4

h(x) = x2 − 2x − 3 5 0 −3 −4 −3 0 5

See Fig. M4.3.4.

6.x −2 −1 0 1 2 3

G(x) = 1 − 2−x −3 −1 0 1/2 3/4 7/8

See Fig. M4.3.6.



y

−4

−3

1

2

3

4

5

6

x−2 1 2 3 4

y

-4-3-2-1

123456

x-2 -1 1 2 3

Figure M4.3.4 Figure M4.3.6

4.4

2. See Figs. M4.4.2a, M4.4.2b, M4.4.2c

y

1

2

3

4

x1 2 3 4

y

−5

−4

−3

−2

−1

1

x1 2 3 4 5 6 7 8 10

y

−1

1

2

3

4

x1 2 3 4 5

Figure M4.4.2a Figure M4.4.2b Figure M4.4.2c

4. L1 is y = x + 2, with slope 1; L2 is y = − 35x + 3, with slope −3/5; L3 is y = 1, with slope 0;

L4 is y = 3x − 14, with slope 3; L5 is y = 19x + 2, with slope 1/9.

6. If P is the price of Q copies, then applying the point–point formula gives P −1400 = 3000−1400500−100 (Q−100)

or P = 1000 + 4Q. The price of printing 300 copies is therefore P = 1000 + 4 · 300 = 2200.

8. (a) See Figs. M4.4.8a, M4.4.8b, and M4.4.8c.

y

x

1

1

y

x−1 1

y

x

1

1




10. See Fig. M4.4.10. Each arrow points to the side of the line where the relevant inequality is satisfied. Theshaded triangle is the required set.

y

x

1

2

3

2 3 4

3x + 4y = 123x + y = 3

x − y = 1

y

-4-3-2-1

123

x-4 -3 -2 -1 1 2 3

f (x) = − 12 x2 − x + 3

2


4.52. (a) 75 − 3P e = 20 + 2P e, and hence P e = 11. (b) P e = 90

4. C = 0.8y+100. (The equation is C = ay+b. We are told that 900 = 1000a+b and a = 80/100 = 0.8.So b = 100.)

6. (a) April 1960 corresponds to t = 9/4, when N(9/4) = −17 400 · (9/4) + 151 000 = 111 850.(b) −17 400 t + 151 000 = 0 implies t = 8.68, which corresponds roughly to September 1966.

4.6

2. (a)x −4 −3 −2 −1 0 1 2

f (x) = − 12x2 − x + 3

2 −2.5 0 1.5 2 1.5 0 −2.5

(b) See Fig. M4.6.2. (c) f (x) = − 12 (x + 1)2 + 2. Maximum point (−1, 2).

(d) x = −3 and x = 1. (e) f (x) > 0 in (−3, 1), f (x) < 0 for x < −3 and for x > 1.

4. (a) x(x + 4). Zeros 0 and −4. (b) No factoring is possible. No zeros.(c) −3(x − x1)(x − x2), where the zeros are x1 = 5 + √

15 and x2 = 5 − √15.

(d) 9(x − x1)(x − x2), where the zeros are x1 = 1/3 + √5 and x2 = 1/3 − √

5.(e) −(x + 300)(x − 100). Zeros −300 and 100. (f) (x + 200)(x − 100). Zeros −200 and 100.

6. U(x) has maximum for x = 4(r − 1)/(1 + r2). (Use (5).)

8. (a) If PE = α1 − 13Q denotes the selling price in England, and PG = α2 + 1

6Q denotes the buyingprice in Ghana, π(Q) = (PE − PG − γ )Q = − 1

2Q2 + (α1 − α2 − γ )Q. (b) Using (5), we see thatQ∗ = α1 − α2 − γ maximizes profit if α1 − α2 − γ > 0. If α1 − α2 − γ ≤ 0, then Q∗ = 0.(c) π(Q) = − 1

2Q2 + (α1 − α2 − γ − t)Q and Q∗ = α1 − α2 − γ − t if α1 − α2 − γ − t > 0.(d) T = tQ∗ = t (α1 − α2 − γ − t). Here t = 1

2 (α1 − α2 − γ ) maximizes export tax revenue.

4.72. (a) 1 and −2 (b) 1, 5, and −5 (c) −1

4. (a) y = 12 (x + 1)(x − 3) (b) y = −2(x + 3)(x − 1)(x − 2) (c) y = 1

2 (x + 3)(x − 2)2

6. c4 + 3c2 + 5 = (c2 + 3/2)2 + 11/4 is not 0 for any choice of c, so the division has to leave a remainder.



4.8

2. (a): C (b): D (c): E (d): B (e): A (f) F: y = 2 − (1/2)x

4. (a) 23 = 8, so x = 3/2 (b) 3x + 1 = −4, so x = −5/3 (c) x2 − 2x + 2 = 2, so x2 − 2x = 0,implying that x = 0 or x = 2.

6. V = (4/3)πr3 implies r3 = 3V/4π and so r = (3V/4π)1/3. Hence, S = 4πr2 = 4π(3V/4π

)2/3 =3√

36π V 2/3.

4.9

2. P(t) = 1.22 ·1.034t . The doubling time t∗ is given by the equation (1.034)t∗ = 2, and we find t∗ ≈ 20.7

(years).

4. The graphs are drawn in Fig. M4.9.4.

x −3 −2 −1 0 1 2 3

2x 1/8 1/4 1/2 1 2 4 8

2−x 8 4 2 1 1/2 1/4 1/8

y

2

4

6

8

x−3 −2 −1 1 2 3

y = 2xy = 2−x

y

1

2

x−4 −3 −2 −1 1 2 3 4 5

f (x) = 1 + 4x

x2+4

Figure M4.9.4 Figure M4.R.2

6. We find (1.035)t = 3.91 · 105/5.1 ≈ 0.7667 · 105, and (taking ln of each side) we find t ≈ 327. So theyear is 1969 + 327 = 2296. This is when every Zimbabwean would have only 1 m2 of land on average.

8. (b) and (d) do not define exponential functions. (In (f): y = (1/2)x .)

10. Use the two points on each graph to determine A and b from the equation y = Abx .You then get: (a) y = 2 · 2x (b) y = 2 · 3x (c) y = 4(1/2)x .

4.10

2. (a) ln 3x = x ln 3 = ln 8, so x = ln 8/ ln 3. (b) x = e3 (c) x2 − 4x + 5 = 1 so (x − 2)2 = 0. Hence,x = 2. (d) x(x − 2) = 1 or x2 − 2x − 1 = 0, so x = 1 ± √

2. (e) x = 0 or ln(x + 3) = 0, so x = 0or x = −2. (f)

√x − 5 = 1 so x = 36.



4. (a) t = 1

r − sln

B

A(b) t = 1

0.07ln

5.6

1.2≈ 22. It will take approximately 22 years before the two

countries have the same GNP.

6. (a) Wrong. (Let A = e.) (b) and (c) are right.


2. (a) F(0) = 1, F(−2) = 0, F(2) = 2, and F(3) = 25/13 (b) F(x) = 1 + 4

x + 4/xtends to 1 as x

becomes large positive or negative. (c) See Fig. M4.R.2.

4. (a) |x| ≥ 1, i.e. x ≤ −1 or x ≥ 1 (b) x > 4 (c) 3 ≤ x ≤ 5

6. (a) Slope −4 (b) Slope −3/4 (c) Slope −b/a

8. f (2) = 3 and f (−1) = −3 give 2a+b = 3 and −a+b = −3, so a = 2, b = −1. Hence f (x) = 2x −1and f (−3) = −7. (Or use the point–point formula.)

10. y = 2x2 + x − 6. (−3 = a + b + c, −6 = c, and 15 = 9a + 3b + c)

12. The new profit is πt = 100Q − 52Q2 − tQ, which is maximized at Qt = 1

5 (100 − t).

14. (a) p(x) = x(x − 3)(x + 4) (b) q(x) = 2(x − 2)(x + 4)(x − 1/2)

16. (a) k = 4 (b) k = −3/2 and k = 1 (c) k = 26 (d) k = −1 and k = 4

18. (1 + p/100)15 = 2 gives p = 100(21/15 − 1) ≈ 4.7 as the percentage rate.

20. (a) t = (ln x − b)/a (b) t = (ln 2)/a (c) t = ±√

ln(8/

√2π

)


C H A P T E R 5 P R O P E R T I E S O F F U N C T I O N S 17

Chapter 5 Properties of FunctionsThis chapter begins in Section 5.1 by examining how changes in a function relate to shifts in its graph. Aneconomic application is to shifting demand and supply curves. Section 5.2 considers how to construct newfunctions from old ones, and includes a discussion of symmetry.

Next, Section 5.3 introduces inverse functions. By definition, if a function is one-to-one, there is aninverse from its range or image space back to its domain. The obvious geometric properties of the graphs ofa function and its inverse are explained.

Any equation in two variables can be represented by a curve (or a set of points) in the xy-plane. Someexamples are given in Section 5.4. Also included is the “vertical line test” for a graph to represent a function,and the effect on a graph of changing units. The section ends with some examples of compound functions.

The distance between points in a plane is defined in Section 5.5, and used to derive the equations whosegraph is a circle with a given centre and radius. Ellipses and other conics are briefly discussed.

Section 5.6 concludes the chapter with a discussion of the general concept of a function and the associatedterminology—domain, target, range, image. Definitions of one-to-one functions and of inverse functions arealso given.


5.12. (a) The graph of y = f (x) is moved 2 units to the right. See Fig. M5.1.2a. (b) The graph of y = f (x)

is moved downwards by 2 units. See Fig. M5.1.2b. (c) The graph of y = f (x) is reflected about they-axis. See Fig. M5.1.2c.

y

x

1

1

y

x

1

1

y

x

1

1


4. Move y = |x| two units to the left. Then reflect the graph about the x-axis. Finally, move the graph up 2units. See Fig. M5.1.4. y

x

1

−1

Figure M5.1.4

6. y∗ = Bd2 − Ad + c

2Bd


18 C H A P T E R 5 P R O P E R T I E S O F F U N C T I O N S

5.22. See Figs. M5.2.2a to M5.2.2c. (Note that the graph of e−x2 + x is not symmetric about (0, 1).)

y

x

y

x

y

x


4. f (f (x)) = f (3x + 7) = 3(3x + 7) + 7 = 9x + 28. Then 9x + 28 = 100 for x = 8.

5.32. p = (157.8/D)10/3

4. (a) The domain of f −1 is {−4, −2, 0, 2, 4, 6, 8}. f −1(2) = −1 (b) f (x) = 2x + 4, f −1(x) = 12x − 2

6. (a) f (x) = x/2 and g(x) = 2x are inverse functions. (b) f (x) = 3x − 2 and g(x) = 13 (x + 2) are

inverse functions. (c) C = 59 (F − 32) and F = 9

5C + 32 are inverse functions.

8. (a) See Fig. M5.3.8a. (b) See Fig. M5.3.8b. Triangles OBA and OBC are congruent.

y

x

(3, 1)

(5, 3)(1, 3)

(3, 5)

y = x

y

x

C = (b, a)

A = (a, b)

B

y = x

D

E

y

1

2

3

x1 2 3

Figure M5.3.8a Figure M5.3.8b Figure M5.4.2

10. (a) x = ln y − 4, y > 0 (b) x = ey+4, y ∈ (−∞, ∞) (c) x = 3 + ln(ey − 2), y > ln 2

5.42. (a) See Fig. M5.4.2. (b) If a > 1, then g(a) = 1 − √

a − 1 and g(a) < 1 (see the figure), so thatg(g(a)) = g(1−√

a − 1) = (1−√a − 1)2 −2(1−√

a − 1)+2 = a. If a < 1, then g(a) = a2 −2a+2,and g(a) > 1, so that g(g(a)) = 1−√

a2 − 2a + 2 − 1 = 1−√(a − 1)2 = 1+ (a−1) = a. Moreover,

g(g(1)) = 1. It follows that g(g(a)) = a for all a, so g is its own inverse. (Alternatively, show that onecan solve y = g(x) for x in terms of y, in which case x = g(y).)


C H A P T E R 5 P R O P E R T I E S O F F U N C T I O N S 19

4. F(100 000) = 4070. The graph is the thick line sketched in Fig. M5.4.4.

Y

RN

4070

7500 100 000

y

x5

(2, 4)

2

2

y

x

A = (3, 2)

B = (5, −4)

2

2

P

Figure M5.4.4 Figure M5.5.2 Figure M5.R.10

5.5

2. (5 − 2)2 + (y − 4)2 = 13, or y2 − 8y + 12 = 0, with solutions y = 2 and y = 6. Geometric explanation:The circle with centre at (2, 4) and radius

√13 intersects the line x = 5 at two points. See Fig. M5.5.2.

4. (a) (x − 2)2 + (y − 3)2 = 16 (b) (x − 2)2 + (y − 5)2 = 13

6. The condition is that√

(x + 2)2 + y2 = 2√

(x − 4)2 + y2, which reduces to (x − 6)2 + y2 = 42.

5.6

2. The function in (b) is one-to-one and has an inverse: the rule mapping each youngest child alive today tohis/her mother. (Though the youngest child of a mother with several children will have been different atdifferent dates.) The function in (d) is one-to-one and has an inverse: the rule mapping the surface area tothe volume. The function in (e) is one-to-one and has an inverse: the rule that maps (u, v) to (u − 3, v).The other functions are many-to-one, in general, and so have no inverses.


2. (f + g)(x) = x2 − 2, (f − g)(x) = 2x3 − x2 − 2, (fg)(x) = x2(1 − x)(x3 − 2), (f/g)(x) =(x3 − 2)/x2(1 − x), f (g(1)) = f (0) = −2, and g(f (1)) = g(−1) = 2.

4. p = (64 − 10D)/3

6. (a) x = 50 − 12y (b) x = 5

√y/2 (c) x = 1

3 [2 + ln(y/5)]

8. (a)√

13 (b)√

17 (c)√

(2 − 3a)2 = |2−3a| (Note that 2−3a is the correct answer only if 2−3a ≥ 0,i.e. a ≤ 2/3. Ask students with the wrong answer to put, say, a = 3.)

10. (x − 3)2 + (y − 2)2 = (x − 5)2 + (y + 4)2, which reduces to x − 3y = 7. See Fig. M5.R.10.


20 C H A P T E R 6 D I F F E R E N T I A T I O N

Chapter 6 Differentiation

Our treatment of the single variable calculus begins in Chapter 6. Section 6.1 is devoted to the slopes of curvesand their tangents, as well as the tangent definition of derivative. Section 6.2 discusses how the tangent isthe limit of a sequence of lines joining nearby points on the graph of a function. It proceeds to present theformula for the Newton (or differential) quotient, whose limit is the derivative. The derivative is then used inthe general equation for the tangent to the graph of a function at any point where it is differentiable. A generalrecipe for finding derivatives is provided, and Leibniz’s differential notation is introduced.

Section 6.3 discusses increasing and decreasing functions and explains how the sign of the derivativecan help us to find the intervals of increase/decrease of a differentiable function. (Note 8.4.2 shows how themean-value theorem can be used to prove the main results of this section.)

Section 6.4 is devoted to the economic significance of the derivative of a function. The instantaneousand proportional rates of change are distinguished. Marginal cost, revenue, profit, propensity to consume,product, are all defined in terms of derivatives. Marginal cost is distinguished from incremental cost.

It is crucially important for economics students to understand the definition of the derivative. In fact, thisis more important than knowing how to differentiate very complicated functions. Example 6.4.3 and problemslike 6.4.5 should be emphasized. The procedure of smoothing a function to fit a set of discrete data points isillustrated at the end of this section.

Limits were used to define derivatives informally in Section 6.2. The more careful discussion that isreally needed is the topic of Section 6.5, though even this remains rather informal. The discussion proceedsmostly through examples. Some general rules for limits of sums, differences, products, ratios, and powers arepresented.

After these preliminaries, Section 6.6 is devoted to some simple rules for differentiation, including therule for differentiating a power function. Section 6.7 presents the rules for differentiating sums, differences,products, and quotients. The product rule in particular is illustrated economically. The quotient rule is usedto show that average cost is increasing iff it is less than marginal cost.

Section 6.8 states the chain rule, first using Leibniz’s notation in a suggestive way, then more carefully.An incomplete proof of the chain rule is provided in small print; we judge that a complete proof belongs tomore rigorous courses in mathematical analysis.

The next section 6.9 introduces second-order derivatives. It also defines convex and concave functionsusing the sign of the second derivative. In fact, much modern economic analysis has been built on specificassumptions to the effect that certain functions are either concave or convex. The examples illustrated inFigures 4 and 5, or similar examples, should be carefully discussed. In Example 5 it is shown that an increas-ing concave transformation of a concave function is concave. Higher-order derivatives, found by repeateddifferentiation, are discussed next.

Sections 4.9 and 4.10 introduced exponential and logarithmic functions, which are often used in econom-ics, of course, as well as in statistics. Section 6.10 shows that if the special number e = 2.718281828 . . . isused as the base for the exponential function, then dex/dx = ex for all x, so the derivative of the function is thefunction itself. It also shows how to differentiate eg(x) using the chain rule. Finally, the rule for differentiatingthe general exponential function ax is derived.

Section 6.11 shows how to differentiate the natural logarithmic function, d ln x/dx = 1/x, and alsothe derivative d ln[h(x)]/dx for any positive valued differentiable function h(x). The section includes adiscussion of the useful technique of logarithmic differentiation—i.e., taking the logarithm of an expressionbefore differentiating. As shown later in Section 7.7, this is often a natural way for economists to calculateelasticities.


C H A P T E R 6 D I F F E R E N T I A T I O N 21

A subsection characterizes the number e as limh→0(1 + h)1/h. Finally, when a is any real number andx > 0, the power xa can be defined as exp(a ln x), and using the chain rule, the power rule then follows forall real exponents a.

In fact, Chapter 6 almost completes the inventory of functions of a single variable used in this book, and inmost mathematical work in economics. The major omission is the family of trigonometric functions discussedin FMEA. In economics, they are used almost exclusively to solve difference and differential equations, whichare topics in FMEA.


6.1

2. g(5) = 1, g′(5) = 1

6.2

2. (a) f ′(x) = 6x + 2 (b) f ′(0) = 2, f ′(−2) = −10, f ′(3) = 20. The tangent equation is y = 2x − 1.

4.f (x + h) − f (x)

h= 1/(x + h) − 1/x

h= x − (x + h)

hx(x + h)= −h

hx(x + h)= −1

x(x + h)−→h→0

− 1

x2

6. (a) f (x + h) − f (x) = a(x + h)2 + b(x + h) + c − (ax2 + bx + c) = 2ahx + bh + ah2, so[f (x + h) − f (x)]/h = 2ax + b + ah → 2ax + b as h → 0. Thus f ′(x) = 2ax + b.(b) f ′(x) = 0 for x = −b/2a. The tangent is parallel to the x-axis at the minimum/maximum point.

8. (a) Use the difference of squares formula. (b) [f (x +h)−f (x)]/h = (√x + h−√

x)/h. The identityin (a) yields the result. (c) Letting h → 0, the formula follows. (The students may need reminding that√

x = x1/2 and 1/√

x = x−1/2.)

10.f (x + h) − f (x)

h= (x + h)1/3 − x1/3

h= 1

(x + h)2/3 + (x + h)1/3x1/3 + x2/3→ 1

3x2/3as h → 0,

and 1/(3x2/3) = 13x−2/3.

6.3

2. f ′(x) = −3x2 +8x −1 = −3(x −x0)(x −x1), where x0 = 13 (4−√

13) ≈ 0.13 and x1 = 13 (4+√

13) ≈2.54. Then f (x) is increasing in [x0, x1], decreasing in (−∞, x0] and in [x1, −∞).

6.4

2. I is the fixed cost, whereas k is the marginal cost, and also the (constant) incremental cost of producingeach additional unit.

4. T ′(y) = t , so the marginal tax rate is constant.

6. (a) C ′(x) = 3x2 − 180x + 7500 (b) x = 30 (C ′(x) has a minimum at x = 180/6 = 30, using (4.6.5).)

6.5

2. (a) 0.6931 (b) 1.0986 (c) 0.4055 (Actually, using the result in Example 7.12.2, the precise values ofthese limits are ln 2, ln 3, and ln(3/2), respectively.)


22 C H A P T E R 6 D I F F E R E N T I A T I O N

4. (a) 5 (b) 1/5 (c) 1 (d) −2 (e) 3x2 (f) h2

6. (a) 4 (= f ′(1)) (b) 5 (= f (2) − f (1)) (c) 6 (= f ′(2)) (d) 2a + 2 (= f ′(a)) (e) 2a + 2 (= f ′(a))(f) [f (a+h)−f (a−h)]/h = [(a+h)2+2(a+h)−(a−h)2−2(a−h)]/h = 4a+4 → 4a+4 as h → 0, sof ′(a) = 4a+4. (Alternatively, [f (a+h)−f (a−h)]/h = [f (a+h)−f (a)]/h+[f (a)−f (a−h)]/h →f ′(a) + f ′(a) = 4a + 4 as h → 0.)

6.6

2. (a) 2g′(x) (b) − 16g′(x) (c) 1

3g′(x)

4. (a) 8πr (b) A(b + 1)yb (c) (−5/2)A−7/2

6. (a) F(x) = 13x3 + C (b) F(x) = x2 + 3x + C (c) F(x) = xa+1

a + 1+ C, where in all three cases C is

an arbitrary constant.

6.7

2. (a) 65x − 14x6 − 1

2x−1/2 (b) 4x(3x4 − x2 − 1) (c) 10x9 + 5x4 + 4x3 − x−2

4. (a)3

2√

x(√

x + 1)2(b)

−x4 + 5x2 + 18x + 2

(x2 + 2)2(x + 3)2(c)

−2(1 + 2x)

x3(d)

4x

(x2 + 1)2(e)

−2x2 + 2

(x2 − x + 1)2

(f)2(x3 − x2 + x + 1)

3x3(x + 1)2

6. (a) [2, ∞) (b)[−√

3, 0]

and in[√

3, ∞)(c)

[−√2,

√2]

(d) (−∞, x1] and in [0, x2], where x1,2 = − 12 ∓ 1

2

√5. (f ′(x) = −x(x − x1)(x − x2)

(x + 1)2)

8. f ′(x) = 3(x − 1)(x + 1)

(−x2 + 4x − 1)2. f (x) is increasing in (−∞, −1], in

[1, 2 + √

3), and in

(2 + √

3, ∞).

10. Differentiating f (x) · f (x) = x gives f ′(x) · f (x) + f (x) · f ′(x) = 1, so 2f ′(x) · f (x) = 1. Hence,

f ′(x) = 1

2f (x)= 1

2√

x.

6.8

2. (a) dY/dt = (dY/dV )(dV/dt) = (−3)5(V + 1)4t2 = −15t2(t3/3 + 1)4

(b) dK/dt = (dK/dL)(dL/dt) = AaLa−1b = Aab(bt + c)a−1

4. (dY/dt)t=t0 = (dY/dK)t=t0 · (dK/dt)t=t0 = Y ′(K(t0))K′(t0) 6.

dx

dp= − a

2√

ap − c

8. b(t) is the total fuel consumption after t hours. b′(t) = B ′(s(t))s ′(t). So the rate of fuel consumptionper hour is equal to the rate per kilometre multiplied by the speed in kph.

10. (a) y ′ = 5(x4)4 · 4x3 = 20x19 (b) y ′ = 3(1 − x)2(−1) = −3 + 6x − 3x2

12. (a) 1 + f ′(x) (b) 2f (x)f ′(x) − 1 (c) 4[f (x)

]3f ′(x) (d) 2xf (x) + x2f ′(x) + 3

[f (x)

]2f ′(x)

(e) f (x) + xf ′(x) (f)f ′(x)

2√

f (x)(g)

2xf (x) − x2f ′(x)

[f (x)]2(h)

2xf (x)f ′(x) − 3[f (x)]2

x4


C H A P T E R 6 D I F F E R E N T I A T I O N 23

6.92. d2y/dx2 = (1 + x2)−1/2 − x2(1 + x2)−3/2 = (1 + x2)−3/2 4. g′′(2) = 2. (g′′(t) = 2(t − 1)−3)

6. d2L/dt2 = 3(2t − 1)−5/2 8. Because g(u) is not concave. 10. d3L/dt3 > 0

6.10

2. (a) x ′ = et [a + b + t (b + 2c + ct)] (b) x ′ = −qt4 + 2qt3 − pt − p

t2et

(c) x ′ = 2(at + bt2)(a + 2bt)et − (at + bt2)2et

(et )2= t (a + bt)(−bt2 + (4b − a)t + 2a)

et

4. (a) (−∞, ∞) (b) [0, 1/2] (c) (−∞, −√2/2] and in [0,

√2/2]

6. (a) eex

ex = eex+x (b) 12 (et/2 − e−t/2) (c) − et − e−t

(et + e−t )2(d) z2ez3

(ez3 − 1)−2/3

6.11

2. (a) x2 ln x(3 ln x + 2) (b)x(2 ln x − 1)

(ln x)2(c)

10(ln x)9

x(d)

2 ln x

x+ 6 ln x + 18x + 6

4. (a) x > −1 (b) 1/3 < x < 1 (c) x �= 0

6. (a) (−2, 0] (y is defined only in (−2, 2).) (b) [e−1/3, ∞) (y ′ = x2(3 ln x + 1), x > 0)(c) [e, e3] (y ′ = (1 − ln x)(ln x − 3)/2x2, x > 0)

8. (a)f ′(x)

f (x)= −2

3(x2 − 1)(b)

f ′(x)

f (x)= 2 ln x + 2 (c)

f ′(x)

f (x)= 1

2x − 4+ 2x

x2 + 1+ 4x3

x4 + 6

10. ln y = v ln u, so y ′/y = v′ ln u + vu′/u.


2. [f (x + h) − f (x)]/h = −6x2 + 2x − 6xh − 2h2 + h → −6x2 + 2x as h → 0, so f ′(x) = −6x2 + 2x.

4. Because C ′(1000) ≈ C(1001) − C(1000), if C ′(1000) = 25, the additional cost of producing 1 morethan 1000 units is approximately 25. It is profitable to increase production if each unit is sold for 30.

6. Because A′(100) ≈ A(101) − A(100), the additional cost of increasing the area from 100 to 101 m2 isapproximately 250 dollars.

8. (a) 2at (b) a2 − 2t (c) 2xφ − 1/2√

φ

10. (a) dZ/dt = (dZ/du)(du/dt) = 3(u2 − 1)22u3t2 = 18t5(t6 − 1)2

(b) dK/dt = (dK/dL)(dL/dt) = (1/2√

L)(−1/t2) = −1/2t2√1 + 1/t

12. dR/dt = (dR/dS)(dS/dK)(dK/dt) = αSα−1βγKγ−1Aptp−1 = Aαβγptp−1Sα−1Kγ−1

14. (a) y ′ = −7ex (b) y ′ = −6xe−3x2(c) y ′ = xe−x(2 − x) (d) y ′ = ex[ln(x2 + 2) + 2x/(x2 + 2)]

(e) y ′ = 15x2e5x3(f) y ′ = x3e−x(x − 4) (g) y ′ = 10(ex + 2x)(ex + x2)9

(h) y ′ = 1/2√

x(√

x + 1)

16. (a) dπ/dQ = P(Q) + QP ′(Q) − c (b) dπ/dL = PF ′(L) − w


24 C H A P T E R 7 D E R I V A T I V E S I N U S E

Chapter 7 Derivatives in Use

Implicit differentiation, often used in economic analysis, is the subject of the first three sections. Section7.1 concentrates on theory, with purely mathematical examples and problems, while in Section 7.2 severalimportant economic examples are discussed. In Section 7.3 the formula for the derivative of an inverse functionis derived.

Linear approximations and differentials are treated in Section 7.4. It is important for students to learnhow differentials are manipulated, since economists often make use of them.

Then quadratic and general polynomial approximations are discussed in Section 7.5, and the nth orderTaylor polynomial is introduced.

Section 7.6 extends the discussion of Section 7.5 and introduces the Lagrange remainder formula. Thisleads to the Taylor formula which is an important result in mathematical analysis.

Section 7.7 introduces elasticities, and their definition in terms of derivatives. The alternative character-ization of elasticities as logarithmic derivatives is explained.

Continuity is the subject of Section 7.8. After some preliminary discussion, the need to consider continu-ity is explained, then a formal definition in terms of limits is provided. It is pointed out that sums, differences,products, powers, and composites of continuous functions are continuous. So are quotients, where the de-nominator is not zero.

Section 7.9 continues the discussion of limits that was started in Section 6.5. In some sense, limits andcontinuity are logically prior to the first results on differentiation contained in Chapter 6. Nevertheless, wehave chosen this order of presentation because the topics treated in Section 7.9 may seem rather abstractunless one has first seen limits being used in connection with derivatives. One-sided limits are discussed, asare one-sided derivatives for functions whose graphs may have corners. Limits at infinity are briefly described,while Problem 7.9.4 introduces the general definition of asymptotes. In this section we also point out thatdifferentiable functions are continuous.

A rigorous definition of limits and continuity is discussed in small print. We feel that at least somestudents can already benefit at this stage from seeing the εδ-definition of limits and continuity, although a realunderstanding of these concepts is beyond this text.

Section 7.10 has a discussion of the intermediate value theorem. Newton’s method for finding the rootsof equations also receives brief treatment.

Section 7.11 deals with infinite sequences and has a short description of irrational numbers as limits ofsequences. The chapter ends with Section 7.12 discussing l’Hôpital’s rule for evaluating limits, which is quiteoften used in economics.


7.1

2. 2u+v+udv

du−3v2 dv

du= 0, so

dv

du= 2u + v

3v2 − u. Hence

dv

du= 0 when v = −2u (provided 3v2 −u �= 0).

Substituting for v in the original equation yields 8u3 − u2 = 0. So the only point on the curve wheredv/du = 0 and u �= 0 is (u, v) = (1/8, −1/4).

4. (a) y ′ = −x/y (b) y ′ = −√y/x (c) y ′ = 2x(2x2 − y3)/y2(3x2 + 4y)

6. y ′ = 1

F ′(0) + 1. (Differentiation w.r.t. x yields 3x2F(xy) + x3F ′(xy)(y + xy ′) + exy(y + xy ′) = 1.

Then put x = 1, y = 0. Note that F is a function of only one variable, with argument xy.)


C H A P T E R 7 D E R I V A T I V E S I N U S E 25

7.2

2. (a) 1 = C ′′(Q∗)(dQ∗/dP ), so dQ∗/dP = 1/C ′′(Q∗) (b) dQ∗/dP > 0, which is reasonable becauseif the price received by the producer increases, the optimal production should increase.

4. (a)Y = f (Y )+I+X−g(Y ). Differentiating w.r.t. I yieldsdY/dI = f ′(Y )(dY/dI)+1−g′(Y )(dY/dI).Thus, dY/dI = 1/

[1 − f ′(Y ) + g′(Y )

]. Imports should increase when income increases, so g′(Y ) > 0.

We find that dY/dI > 0. (b) d2Y/dI 2 = (f ′′−g′′)(dY/dI)/(1−f ′+g′)2 = (f ′′−g′′)/(1−f ′+g′)3.

7.3

2. (a) f ′(x) = 4x2(3 − x2)

3√

4 − x2. So f increases in [−√

3,√

3 ], and decreases in [−2, −√3 ] and in [

√3, 2].

See Figure M7.3.2. (b) f has an inverse in the interval [0,√

3] because f is strictly increasing there.g′( 1

3

√3 ) = 1/f ′(1) = 3

√3/8.

y

−2

−1

1

2

x−2 −1 1 2r

r + dr


7.4

2.1

(5x + 3)2≈ 1

9− 10

27x. (f (0) = 1/9, f ′(x) = −10(5x + 3)−3, so f ′(0) = −10/27.)

4. AKα ≈ A(1 + α(K − 1)

)6. (a) (pxp−1 + qxq−1) dx (b) (p + q)xp+q−1 dx (c) rp(px + q)r−1 dx (d) (pepx + qeqx)dx

8. (a) (i) �y = 0.61, dy = 0.6 (ii) �y = 0.0601, dy = 0.06(b) (i) �y = 0.011494, dy = 0.011111 (ii) �y = 0.001115, dy = 0.001111(c) (i) �y = 0.012461, dy = 0.0125 (ii) �y = 0.002498, dy = 0.0025

10. (a) A(r + dr) − A(r) is the shaded area in Fig. M7.4.10. It is approximately equal to the length of theinner circle, 2πr , times dr .(b) V (r + dr) − V (r) is the volume of the shell between the ball with radius r + dr and the ball withradius r . It is approximately equal to the area of the inner ball 4πr2 times dr .

7.5

2. p(x) = x − 12x2 + 1

3x3 − 14x4 + 1

5x5

4. Follows from formula (1) with f = U , a = y, x = y + M − s.



6. We find x(0) = 2[x(0)]2 = 2. Differentiating the expression for x(t) yields x(t) = x(t) + t x(t) +4[x(t)]x(t), and so x(0) = x(0)+4[x(0)]x(0) = 1+4·1·2 = 9. Hence, x(t) ≈ x(0)+x(0)t+ 1

2 x(0)t2 =1 + 2t + 9

2 t2.

8. Use (2) with f (x) = (1+x)n and x = p/100. Then f ′(x) = n(1+x)n−1 and f ′′(x) = n(n−1)(1+x)n−2.The approximation follows.

7.6

2. (a) 3√

25 = 3(1 − 2/27)1/3 ≈ 3(1 − 13

227 − 1

94

272 ) ≈ 2.924

(b) 5√

33 = 2(1 + 1/32)1/5 ≈ 2(1 + 15·32 − 2

251

322 ) ≈ 2.0125

4. (a) p(x) = 1 + 13x − 1

9x2 (b) |R3(x)| = ∣∣ 13!g

′′′(c)x3∣∣ = ∣∣ 1

61027 (1 + c)−8/3x3

∣∣ ≤ 581x3

(c) 3√

1003 = 10(1+3 ·10−3)1/3 ≈ 10.0099900, using part (a) to approximate (1+3 ·10−3)1/3. The errorin (b) is |R3(x)| ≤ 5

81 (3 · 10−3)3 = 53 10−9. So the error in 3

√1003 is ≤ 10|R3(x)| = 50

3 10−9 < 2 · 10−8,and the answer is correct to 7 decimal places.

7.7

2. ElKT = 1.06. A 1% increase in expenditure on road building leads to an increase in the traffic volumeof approx. 1.06 %.

4. (a) Elxeax = (x/eax)aeax = ax (b) 1/ ln x (c) p + ax (d) p + 1/ ln x

6. ElrD = 1.23. This means that a 1% increase in income leads to an increase in the demand for apples ofapprox. 1.23 %

8. (a) See Fig. M7.7.8. (The straight line has been drawn through the lowest and highest points.)

T 36.3 35.0 33.9 32.4 24.7 24.2

ln n 5.04 4.89 4.70 4.54 3.64 3.58

(b) f (T ) = 1.99e0.12T (c) The fall in temperature that halves the pulse rate is (ln 2)/0.12 ≈ 5.8 degrees.

ln n

T

5

4

3

20 25 30 35 40

Figure M7.7.8

10. (a) ElxA = (x/A)(dA/dx) = 0

(b) Elx(fg) = x

fg(fg)′ = x

fg(f ′g + fg′) = xf ′

f+ xg′

g= Elxf + Elxg



(c) Elxf

g= xg

f

(f

g

)′= xg

f

(gf ′ − fg′

g2

)= xf ′

f− xg′

g= Elxf − Elxg

(d) Elx(f + g) = x(f ′ + g′)f + g

=f

xf ′

f+ g

xg′

g

f + g= f Elxf + gElxg

f + g(e) Is like (d), but with +g replaced by −g, and +g′ by −g′.(f) z = f (g(u)), u = g(x) ⇒ Elxz = x

z

dz

dx= x

u

u

z

dz

du

du

dx= Eluf (u) Elxu

7.82. f is discontinuous at x = 0. g is continuous at x = 2. The graphs of f and g are shown in Figs. M7.8.2a

and M7.8.2b.y

−3

−2

1

2

3

4

x−2 1 2 3

f

y

−2

−1

1

2

3

4

x1 2 3 4 5

g

Figure M7.8.2a Figure M7.8.2b

4. See Fig. M7.8.4; y is discontinuous at x = a.

6. See Fig. M7.8.6. (This example shows that the commonly seen statement: “if the inverse function ex-ists, the original and the inverse function must both be monotonic” is wrong. This claim is correct forcontinuous functions, however.)

y

xa

y

x

y

x

1

1

f (x)

Figure M7.8.4 Figure M7.8.6 Figure M7.9.6

7.9

2. (a)x − 3

x2 + 1=

1

x− 3

x2

1 + 1

x2

→ 0 as x → ∞. (b)

√2 + 3x

x − 1=√

3 + 2/x

1 − 1/x→ √

3 as x → ∞. (c) a2



4. (a) y = x − 1 (x = −1 is a vertical asymptote). (b) y = 2x − 3 (c) y = 3x + 5 (x = 1 is a verticalasymptote). (d) y = 5x (x = 1 is a vertical asymptote).

6. f ′(0+) = 1 and f ′(0−) = 0. See Fig. M7.9.6.

7.102. A person’s height is a continuous function of time (even if growth occurs in intermittent spurts, often

overnight). The intermediate value theorem (and common sense) give the conclusion.

4. Integer root: x = −3. Newton’s method gives −1.880, 0.346, and 1.533 for the three other roots.

6. If f (x0) and f ′(x0) have the same sign (as in Fig. 2), then (1) implies that x1 < x0. But if they haveopposite signs, then x1 > x0.

7.11

2. (a) Converges to 5. (b) Diverges to ∞. (c) Converges to 3√

2/2. (sn = 3n

n√

2 − 1/n2= 3√

2 − 1/n2

→ 3/√

2 = 3√

2/2 as n → ∞)

7.12

2. (a) limx→a

x2 − a2

x − a= “0

0

” = limx→a

2x

1= 2a (or

x2 − a2

x − a= (x − a)(x + a)

x − a= x + a → 2a as x → a.)

(b) limx→0

2(1 + x)1/2 − 2 − x

2(1 + x + x2)1/2 − 2 − x= “0

0

” = limx→0

(1 + x)−1/2 − 1

(1 + 2x)(1 + x + x2)−1/2 − 1= “0

0

” =

limx→0

(−1/2)(1 + x)−3/2

2(1 + x + x2)−1/2 + (−1/2)(1 + 2x)2(1 + x + x2)−3/2= −1/2

2 − 1/2= −1

3

4. The second fraction is not “0/0”. The correct limit is 5/2.

6. limx→∞

f (x)

g(x)= lim

t→0+f (1/t)

g(1/t)= “0

0

” = limt→0+

f ′(1/t)(−1/t2)

g′(1/t)(−1/t2)= lim

t→0+f ′(1/t)

g′(1/t)= lim

x→∞f ′(x)

g′(x)


2. 5y4y ′ − y2 − 2xyy ′ = 0, so y ′ = y2

5y4 − 2xy= y

5y3 − 2xfor y �= 0 and 2x �= 5y3. Because no point

with y = 0 can satisfy the given equation, y ′ is never 0.

4. (a) y ′ = −4/13 (Implicit differentiation yields (∗) 2xy + x2y ′ + 9y2y ′ = 0.)(b) Differentiating (∗) w.r.t. x yields: 2y + 2xy ′ + 2xy ′ +x2y ′′ + 18yy ′y ′ + 9y2y ′′ = 0. Inserting x = 2,y = 1, and y ′ = −4/13 gives the answer.

6. y ′ = −(2/x)(1 + 15 ln x)

1 + 1/y= 0 for x = e−5.

8. (a) f (e2) = 2 and f (x) = ln x(ln x − 1)2 = 0 for ln x = 0 and ln x = 1, so x = 1 or x = e.(b) f ′(x) = (3/x)(ln x − 1)(ln x − 1/3) > 0 for x > e, and so f is strictly increasing in [e, ∞). Ittherefore has an inverse h. According to (7.3.3), because f (e2) = 2, we have h′(2) = 1/f ′(e2) = e2/5.

10. (a) x dx/√

1 + x2 (b) 8πr dr (c) 400K3 dK (d) −3x2 dx/(1 − x3)



12. Letting x = 12 and n = 5, formula (7.6.6) yields,

e12 = 1 +

12

1!+ ( 1

2 )2

2!+ ( 1

2 )3

3!+ ( 1

2 )4

4!+ ( 1

2 )5

5!+ ( 1

2 )6

6!ec

where c is some number between 0 and 12 . Now, R6(

12 ) = ( 1

2 )6

6! ec <( 1

2 )6

6! 2 = 123040 ≈ 0.0004340, where

we used the fact that since c < 12 , ec < e

12 < 2. Thus it follows that

e12 ≈ 1 +

12

1!+ ( 1

2 )2

2!+ ( 1

2 )3

3!+ ( 1

2 )4

4!+ ( 1

2 )5

5!= 1 + 1

2+ 1

8+ 1

48+ 1

384+ 1

3840≈ 1.6486979

The error is less than 0.000043, and e12 ≈ 1.649 correct to 3 decimals.

14. We find y ′ = 1/2 and y ′′ = 1/8, so y(x) ≈ 1 + 12x + 1

16x2.

16. Elr (Dmarg) = −0.165 and Elr (Dmah) = 2.39. When income increased by 1%, the demand for mar-garine decreased by approximately 0.165 %, while the demand for meals away from home increased byapproximately 2.39 %.

18. Put f (x) = x3 − x − 5. Then f ′(x) = 3x2 − 1. Taking x0 = 2, formula (7.10.1) with n = 1 givesx1 = 2 − f (2)/f ′(2) = 1 − 1/11 ≈ 1.909.

20. (a) 2 (b) Approaches +∞. (c)3 − √

x + 17

x + 1approaches +∞ as x → −1−, but −∞ as x → −1+,

so there is no limit as x → −1.

22. (a) 1 (b) −1/16 (c) limx→−1

2ex+1 − x2 − 4x − 5

(x + 1)3= “0

0

” = limx→−1

2ex+1 − 2x − 4

3(x + 1)2= “0

0

” =

limx→−1

2ex+1 − 2

6(x + 1)= “0

0

” = limx→−1

2ex+1

6= 1

3.

24. Does not exist if b �= d . If b = d , the limit is “0/0” and by l’Hôpital’s rule, it islimx→0

[ 12a(ax + b)−1/2 − 1

2c(cx + d)−1/2]/1 = a/2

√b − c/2

√d = (a − c)/2

√b.

26. x1 = 0.9−f (0.9)/f ′(0.9) ≈ 0.9247924, x2 = x1−f (x1)/f′(x1) ≈ 0.9279565, x3 = x2−f (x2)/f

′(x2)

≈ 0.9280338, and x4 = x3−f (x3)/f′(x3) ≈ 0.9280339. So it seems that the answer correct to 3 decimals

is 0.928.


30 C H A P T E R 8 S I N G L E - V A R I A B L E O P T I M I Z A T I O N

Chapter 8 Single-Variable Optimization

The previous chapters have presented most of the mathematical functions used in economic analysis, as wellas rules for differentiating them, etc. After these essential preliminaries, Chapter 8 turns to the systematicstudy of single variable optimization problems.

In Section 8.1 we begin by giving algebraic (non-calculus) definitions of maxima and minima. Then wedefine a stationary point, and discuss the role played by stationary points in optimization theory. Studentsshould be encouraged even more strongly than usual to do at least some of the problems in this section inorder to understand that sometimes extreme points can be found by merely examining the formula for thefunction, without any need for differentiation.

Section 8.2 gives an extremely useful and important test for (global) maxima and minima, based onlyon the variation in the sign of the first derivative (Theorem 8.2.1). When this test is passed, there is noneed whatsoever to consider the second derivative. (We emphasize this point because it is often neglected byeconomists who, in our experience, are routinely taught to consider second-derivative tests even when thereis no need to do so.)

Theorem 8.2.2 claims that, at an interior stationary point of an interval in which a function is concave (orconvex), that function has a maximum (or minimum) over that interval. This is a result of great significancein many economic applications. It is also a result that extends to functions of n variables which are concaveor convex on some convex set. The case n = 2 is covered in Theorem 13.2.1.

Section 8.3 presents some interesting economic examples. The economic interpretation of condition (∗)

in Example 1 should be noted. Example 3 illustrates an important type of sensitivity analysis of an optimizationproblem, and illustrates the envelope theorem in a special case.

Next, in Section 8.4, the extreme-value theorem is formulated. It is then pointed out that extreme points,if they are not stationary points, must be either end points of an interval on which the function is defined, orelse points where the function is not even differentiable. This is the justification for the usual procedure ofsearching for extrema by considering stationary points. But it is important to realize when the procedure failsbecause the relevant extreme point is not an interior point of the domain at which the function is differentiable.Otherwise, one would be restricted to interior and “smooth” maxima for the rest of one’s work in economicanalysis—and, of course, corner solutions are often important. Finally, the mean-value theorem is formulatedand used to confirm the tests for increasing/decreasing functions discussed in Section 6.3.

In Section 8.5 further economic examples are studied. Again, the economic interpretations of the first-order conditions should be stressed.

Section 8.6 moves on to local rather than global extreme points. First, these are defined algebraically(using the concept of a neighbouring interval about a local extreme point). Theorem 8.6.1 gives conditionsfor a stationary point to be a local maximum or a local minimum, or neither, based on the variation in the signof the derivative near the stationary point. (These tests are often erroneously claimed to be not only sufficientbut also necessary for local optimality.)

ThenTheorem 8.6.2 proceeds to the rather standard second-derivative test. This gives a sufficient conditionfor a (strict) local maximum (or minimum), based on the function being twice differentiable around thestationary point, with its second derivative being negative (or positive) at that point. (If f ′′(c) = 0, the testfails. Then Theorem 8.6.1 can usually be applied. Tests based on evaluating higher order derivatives at c, webelieve, are only of marginal interest to economists. Again, these nth order tests are often erroneously claimedto be necessary and sufficient.) As a simple exercise in comparative statics, Example 5 shows how the secondderivative at an optimum can be used to infer the sign of a response to an exogenous shock.


C H A P T E R 8 S I N G L E - V A R I A B L E O P T I M I Z A T I O N 31

Section 8.7 considers inflection points, for which Theorem 8.7.1 gives a simple test. This section alsoprovides an alternative definition of concavity/convexity based on line segments joining pairs of points onthe graph. This leads next to standard definitions of strict concavity or convexity, for which there are obvioussecond-derivative sufficient conditions. These ideas will be generalized and extensively discussed in FMEA.


8.1

2. (a) Minimum −1 at x = 0. (F(0) = −1 and−2

2 + x2≥ −1 for all x because 2 + x2 ≥ 2 and so

2

2 + x2≤ 1.) No maximum. (b) Maximum 2 at x = 1. No minimum.

(c) Minimum 99 at x = 0. No maximum. (When x → ±∞ , H(x) → 100.)

8.2

2. h′(x) = 8(2 − √3x)(2 + √

3x)

(3x2 + 4)2. The function has a maximum at x = 2

√3/3 and a minimum at

x = −2√

3/3, with h(2√

3/3) = 2√

3/3 and h(−2√

3/3) = −2√

3/3.

4. f ′(x) = [4x(x4 + 1) − 2x24x3]/(x4 + 1)2, then simplify and factor. f on [0, ∞) has maximum 1 atx = 1, because f (x) increases in [0, 1] and decreases in [1, ∞).

6. f ′(x) = 3e3x − 6ex = 3ex(e2x − 2). So f ′(x) = 0 when e2x = 2 or 2x = ln 2, so x = 12 ln 2. If

x < 12 ln 2 then f ′(x) < 0, and if x > 1

2 ln 2 then f ′(x) > 0, so x = 12 ln 2 is a minimum point.

f (x) = ex(e2x − 6) tends to +∞ as x → ∞, so f has no maximum.

8. (a) x = 13 ln 2 is a minimum point (y is convex). (b) x = 1

3 (a + 2b) is a maximum point (y is concave).(c) x = 1

5 is a maximum point (y is concave).

10. (a) f ′(x) = k − Aαe−αx = 0 when x0 = (1/α) ln(Aα/k). Note that x0 > 0 iff Aα > k. Moreover,f ′(x) < 0 if x < x0 and f ′(x) > 0 if x > x0, so x0 solves the minimization problem.(b) Substituting for A in the answer to (a) gives the expression for the optimal height x0. Its value increasesas p0 (probability of flooding) or V (cost of flooding) increases, but decreases as δ (interest rate) or k

(marginal construction costs) increases. The signs of these responses are obviously what an economistwould expect.

8.32. (a) π(Q) = (102 − 2Q)Q − (2Q + 1

2Q2) = 100Q − 52Q2, which is maximized at Q = 20.

(b) π(Q) = 100Q − 52Q2 − tQ, which is maximized at Q = 20 − t/5.

4. p′(x) = kce−cx , p′′(x) = −kc2e−cx . No maximum exists, and p(x) → a + k as x → ∞. SeeFig. M8.3.2.

8.42. (a) Maximum −1 at x = 0. Minimum −7 at x = 3.

(b) Maximum 10 at x = −1 and x = 2. Minimum 6 at x = 1.(c) Maximum 5/2 at x = 1/2 and x = 2. Minimum 2 at x = 1.(d) Maximum 4 at x = −1. Minimum −6

√3 at x = √

3.(e) Maximum 4.5 · 109 at x = 3000. Minimum 0 at x = 0.


32 C H A P T E R 8 S I N G L E - V A R I A B L E O P T I M I Z A T I O N

p

x

a + k

p(x) = a + k(1 − e−cx)

a 1000

2000

3000

4000

Q

R(Q) = 80Q

C(Q) = Q2 + 10Q + 900

10 Q0 30 Q∗ 40 50


4. g′(x) = 25xex2

(1 − e2−2x2). Stationary points: x = 0 and x = ±1. Here x = 2 is a maximum point,

x = 1 and x = −1 are minimum points. (Note that g(2) = 15 (e4 + e−2) > g(0) = 1

5 (1 + e2).)

6. (a) x∗ = 3/2 (b) x∗ = √2/2 (c) x∗ = √

12 (d) x∗ = √3

8. f is not continuous at x = −1 and x = 1. It has no maximum because f (x) is arbitrarily close to 1 forx sufficiently close to 1. But there is no value of x for which f (x) = 1. Similarly, there is no minimum.

8.5

2. (a) See Fig. M8.5.2. (b) (i) The requirement is π(Q) ≥ 0 and Q ∈ [0, 50], that is −Q2+70Q−900 ≥ 0and Q ∈ [0, 50]. The firm must produce at least Q0 = 35 − 5

√13 ≈ 17 units. (ii) Profits are maximized

at Q∗ = 35.

4. (i) Q∗ = 450 (ii) Q∗ = 550 (iii) Q∗ = 0

6. π(Q) is stationary at Q = (P/ab)1/(b−1). Moreover, π ′′(Q) = −ab(b − 1)Qb−2 < 0 for all Q > 0, sothis is a maximum point.

8.6

2. (a) No local extreme points. (b) Local maximum 10 at x = −1. Local minimum 6 at x = 1.(c) Local maximum −2 at x = −1. Local minimum 2 at x = 1.(d) Local maximum 6

√3 at x = −√

3. Local minimum −6√

3 at x = √3.

(e) No local maximum point. Local minimum 1/2 at x = 3.(f) Local maximum 2 at x = −2. Local minimum −2 at x = 0.

4. a and d are local minimum points, whereas c is a local maximum point for f . Points b and e are neither.

6. (a) x = −3 is a local (and global) minimum point. No local maximum points. (b) x = 0 is a localminimum point and x = −2/ ln 2 is a local maximum point. (Note that f ′(x) = x2x(2 + x ln 2).)

8.7

2. (a) f ′′(x) = 2x(x2 − 3)(1 + x2)−3, so f is convex in [−√3, 0] and in [

√3, ∞). Inflection points are at

x = −√3, 0, and

√3.

(b) g′′(x) = 4(1 + x)−3 > 0 when x > −1, so g is (strictly) convex in (−1, ∞). No inflection points.(c) h′′(x) = (2 + x)ex , so h is convex in [−2, ∞) and x = −2 is an inflection point.


C H A P T E R 8 S I N G L E - V A R I A B L E O P T I M I Z A T I O N 33

4. (a) For x > 0 one has R = p√

x, C = wx + F , and π(x) = p√

x − wx − F .(b) π ′(x) = p(1/2

√x) − w = 0, or p/2

√x = w. (Marginal cost = price times marginal product.) Then

x = p2/4w2. Moreover, π ′′(x) = − 14px−3/2 < 0 for all x > 0, so profit is maximized over the interval

(0, ∞). When x = p2/4w2, then π = p2/2w −p2/4w −F = p2/4w −F . So this is a profit maximumif F ≤ p2/4w; otherwise, the firm does better not to start up and to choose x = 0.

6. a = −2/5, b = 3/5 (f (−1) = 1 gives −a + b = 1. Moreover, f ′(x) = 3ax2 + 2bx and f ′′(x) =6ax + 2b, so f ′′(1/2) = 0 yields 3a + 2b = 0.)

y

x

y

x

f g

y

−2

−1

1

2

x−1 1 2 3 4 5

Figure M8.7.8 Figure M8.R.4

8. See Fig. M8.7.8. Use definition (2).


2. (a) L∗ = 160, L∗∗ = 120 (b) Q′(120) = Q(120)/120 = 720. In general (see Example 6.7.6),(d/dL)(Q(L)/L) = (1/L)(Q′(L) − Q(L)/L). If L > 0 maximizes output per worker, one must haveQ′(L) = Q(L)/L.

4. (a) f has a (global) maximum 27e−3 at x = 3 and no minimum. f has inflection points at x = 0,x = 3 − √

3, and x = 3 + √3. (b) lim

x→∞ f (x) = 0, limx→−∞ f (x) = −∞. See Fig. M8.R.4.

6. (a) f ′(x) = −2x + 1 − e−x , f ′′(x) = −2 + e−x . f ′(x) is strictly increasing in [−3, − ln 2] and strictlydecreasing in [− ln 2, 3].(b) Note that f ′(−3) = 7 − e3 < 0, f ′(− ln 2) = 2 ln 2 − 1 > 0, and f ′(3) = −5 − e−3 < 0. By theintermediate-value theorem and the strict monotonicity of f ′(x) on both sides of x = − ln 2, it followsthat f ′(x) = 0 has just one solution (say, x0) in (−3, − ln 2), and another (actually x = 0) in (− ln 2, 3).(c) Because f decreases in [−3, x0], increases in [x0, 0], then decreases again in [0, 3], the candidatemaxima are x = −3 and x = 0. Because f (−3) = e3 − 12 > f (0) = 1, the maximum is at x = −3.

8. (a) f ′(x) = 4e4x+8ex−32e−2x , f ′′(x) = 16e4x+8ex+64e−2x (b) f ′(x) = 4e−2x(e3x+4)(e3x−2), sof (x) is increasing in [ 1

3 ln 2, ∞), decreasing in (−∞, 13 ln 2]. f ′′(x) > 0 for all x so f is strictly convex.

(c) 13 ln 2 is a (global) minimum. No maximum exists because f (x) → ∞ as x → ∞.

10. (a) g′(x) = (a − 1)(1 − cax−a) and g′′(x) = a(a − 1)cax−a−1. Moreover, limx→0+ g(x) = ∞ andlimx→∞ g(x) = ∞. (b) gmin = g(c) = a(c − 1). (Look at the sign of the derivative.)(c) Because g(c) = a(c−1) < 0, the intermediate-value theorem and strict monotonicity of g(x) on eachside of x = c imply that there is one root in each of the intervals (0, c) and (c, ∞). But g(1) = ca −1 < 0and g(a/(a − 1)) = ca(a/(a − 1))1−a > 0, so the root that is greater than c must lie in (1, a/(a − 1)).


34 C H A P T E R 9 I N T E G R A T I O N

Chapter 9 Integration

Integration is an important mathematical technique which features occasionally in economic analysis. Indeed,after this chapter, integrals appear only rarely in the rest of the book. However, when we study differentialequations and control theory in FMEA, integrals become important. Despite this, undergraduate economicsstudents do need a thorough understanding of elementary ideas in the theory of integration, not so much forthe occasional economic application they may see, but because integrals play such an enormous role in thestatistical foundations of econometrics.

Section 9.1 introduces an indefinite integral as, in effect, an antiderivative—i.e., a function whose de-rivative is the original function. Indefinite integrals are defined only up to an arbitrary additive “constant ofintegration”. General rules for integration are presented.

Section 9.2 shows how integrals are used to find the area of certain plane regions. In fact, it is shown thatif A(x) denotes the area under the graph of the non-negative valued function f (x) over the interval [a, x], thenA(x) is an indefinite integral of f (x). Students should be made aware that this discovery is not another boringfact, but historically a fundamental breakthrough in the history of science. On this basis the area under thegraphs of complicated functions can often be calculated with ease. We next introduce definite integrals. Theycan be found, of course, as differences in the value of any indefinite integral (or antiderivative), provided thatsuch an indefinite integral is known. The section concludes with a brief consideration of the areas associatedwith graphs of functions that may have negative values.

In Section 9.3 standard properties of definite integrals are explained—e.g., the integral of a sum is thesum of the integrals, etc. The rules for differentiating a definite integral w.r.t. both its upper and lower limitsof integration are given. It is pointed out that any continuous function, at least, can be integrated, and a list of“insoluble” integrals is given. Also, it is briefly explained how the “Newton–Leibniz” integral presented inthe chapter can be extended to Riemann integrals (at least for bounded functions).

After three sections of almost pure mathematics comes Section 9.4, with economic applications. The firstconcerns extraction from an oil well. It brings out the elementary but important idea that changes in stocksare integrals of flows. There follows a rather extended discussion of the cumulative distribution function ofincome, defined as the integral of a density function. It is shown how one can find total and mean incomefor all individuals with income in a certain interval. There is also a discussion of how, when all individuals’demands for a commodity are a known function of their incomes, it is possible to find total demand for thatcommodity once the income distribution is also known. The section concludes with a brief discussion ofconsumer and producer surplus.

Next, Section 9.5 presents the method of integration by parts. This is the counterpart for integrals of theproduct rule for differentiation. It is a tricky technique to master, however, because it is not very obviouswhich “part” of the integral needs to be integrated first. So students should be encouraged to attempt as manyproblems as possible.

The second important technique for economics students to learn is that of integration by substitution, asdiscussed in Section 9.6. This is the counterpart for integration of the chain rule for differentiating a functionof a function.

In Section 9.7 we study integrals over infinite intervals, partly because some important economic modelsinvolve an infinite planning horizon, partly because this generalization is useful in statistics.

The chapter ends with a short introduction to differential equations. The first-order equations for propor-tional growth, growth towards an upper limit, and logistic growth are presented and solved. Of course, FMEAprovides a much more extensive discussion of differential equations.


C H A P T E R 9 I N T E G R A T I O N 35


9.1

2. (a) −e−x + C (b) 4e14 x + C (c) − 3

2e−2x + C (d) (1/ ln 2)2x + C

4. (a) 14 t4 + t2 − 3t + C (b) 1

3 (x − 1)3 + C (c) 13x3 + 1

2x2 − 2x + C

(d) 14 (x + 2)4 + C (e) 1

3e3x − 12e2x + ex + C (f) 1

3x3 − 3x + 4 ln |x| + C

6. (a) and (b): Differentiate the right-hand side and check that you get the integrand. (For (a) see alsoProblem 9.5.4.)

8. The graph of f ′(x) in Fig. 2 can be that of a cubic function, with roots at −3, −1, and 1, and withf ′(0) = −1. So f ′(x) = 1

3 (x + 3)(x + 1)(x − 1) = 13x3 + x2 − 1

3x − 1. If f (0) = 0, integrating givesf (x) = 1

12x4 + 13x3 − 1

6x2 − x. Figure M9.1.8 is the graph of this f .

10. (a) Differentiate the right-hand side. (Once we have learned integration by substitution in Section 9.6,this is an easy problem.) (b) (i) 1

10 (2x + 1)5 + C (ii) 23 (x + 2)3/2 + C (iii) −2

√4 − x + C

12. The general form for f ′ is f ′(x) = 13x3 + A, so that for f is f (x) = 1

12x4 + Ax + B. If we require thatf (0) = 1 and f ′(0) = −1, then B = 1 and A = −1, so f (x) = 1

12x4 − x + 1.

y

−1

1

2

x−4 −3 −2 −1 1 2

f

y

x−1 1

f (x) = ex

y

−1

1

2

x−1 1 2


9.2

2. (a)∫ 2

0 3x2 dx = ∣∣20x

3 = 8 (b) 1/7 (c) e − 1/e. (See the shaded area in Fig. M9.2.2.) (d) 9/10

4. A = 12

∫ 1−1(e

x + e−x) dx = 12

∣∣1−1(ex − e−x) = e − e−1

6. (a) f ′(x) = 3x2 − 6x + 2. With x0 = 1 − 13

√3 and x1 = 1 + 1

3

√3, f (x) increases in (−∞, x0) and in

(x1, ∞). (b) See Fig. M9.2.6. The shaded area is∫ 1

0 f (x) dx = ∣∣10(

14x4 − x3 + x2) = 1

4 .

8. (a) 6/5 (b) 26/3 (c) α(eβ − 1)/β (d) − ln 2

9.3

2. (a)1

p + q + 1+ 1

p + r + 1(b) f (x) = 4x3 − 3x2 + 5. (f ′(1) = 6 gives a + b = 6, f ′′(1) = 18 gives

2a + b = 18. Then a = 12 and b = −6, so f ′(x) = 12x2 − 6x. By integration, f (x) = 4x3 − 3x2 + C.Requiring

∫ 20 (4x3 − 3x2 + C) dx = 18 gives C = 5.)



4. (a) Use formula (6) to obtain F ′(x) = x2 + 2. To find G′(x), put u = x2 and use formula (8). ThenG′(x) = [(x2)2 + 2]2x = 2x5 + 4x. (One can also evaluate the integrals and then differentiate: G(x) =∣∣x2

0 ( 13 t3 + 2t) = 1

3x6 + 2x2, so G′(x) = 2x5 + 4x.) (b) H ′(t) = 2tK(t2)e−ρt2(use formula (8)).

6. The parabolas intersect at (0, 0) and (3, 3), so A = ∫ 30 (

√3x − x2 + 2x) dx = 6. See Fig. M9.3.6.

8. (a) f ′(x) = 2√x + 4 (

√x + 4 − 2)

> 0 for x > 0 and f has range (−∞, ∞), so f has an inverse g

defined on (−∞, ∞). We find that the inverse is g(x) = ex/2 + 4ex/4.(b) See Fig. M9.3.8. (c) In Fig. M9.3.8 the graphs of f and g are symmetric about the line y = x, soarea A = area B = 10a − ∫ a

0 (ex/2 + 4ex/4) dx = 10a + 18 − 2ea/2 − 16ea/4. Because a = f (10) =4 ln(

√14 − 2), this simplifies to 10a + 14 − 8

√14 ≈ 6.26.

y

x

y + 1 = (x − 1)2

y2 = 3x

−1

1

2

3

4

−1 1 3 4 5 6

y

−5

5

10

x−5 5 10

B

A a

a

f

g

y = x

50

100

150

t2 4 6 8 10

fg


9.4

2. (a) m = 2b ln 2. (The relevant number of individuals is n∫ 2b

bBr−2 dr = nB/2b, whose total income is

M = n∫ 2b

bBr−1 dr = nB ln 2.) (b) x(p) = ∫ 2b

bnApγ rδBr−2 dr = nABpγ bδ−1(2δ−1 − 1)/(δ − 1)

4. (a) See Fig. M9.4.4. (f and g both have maximum value 4000/27 ≈ 148 at t = 20/3 and at t = 10/3,

respectively.) (b)∫ t

0

(g(τ) − f (τ)

)dτ = 1

2t2(t − 10)2 ≥ 0 for all t .

(c)∫ 10

0p(t)f (t) dt =

∫ 10

0

(−t3 + 9t2 + 11t − 11 + 11/(t + 1))

dt = 940 + 11 ln 11 ≈ 966.38,∫ 10

0p(t)g(t)dt =

∫ 10

0

(t3 − 19t2 + 79t + 121 − 121/(t + 1)

)dt = 3980/3 − 121 ln 11 ≈ 1036.52.

Profile g should be chosen.

6. Equilibrium when6000

Q∗ + 50= Q∗ + 10. The only positive solution is Q∗ = 50, and then P ∗ = 60.

CS =∫ 50

0

[6000

Q + 50− 60

]dQ =

50

0[6000 ln(Q + 50) − 60Q] = 6000 ln 2 − 3000,

PS =∫ 50

0(50 − Q) dQ = 1250



9.5

2. (a)∫ 1

−1x ln(x + 2) dx =

1

−1

12x2 ln(x + 2) −

∫ 1

−1

12x2 1

x + 2dx = 1

2 ln 3 − 12

∫ 1

−1

x2

x + 2dx =

12 ln 3 − 1

2

∫ 1

−1

(x − 2 + 4

x + 2

)dx = 2 − 3

2 ln 3 (b) 8/(ln 2) − 3/(ln 2)2 (c) e − 2

4. Use (1) with f (x) = ln x and g′(x) = xρ . (Alternatively, simply differentiate the right-hand side.)

9.62. (a) 1

24 (2x2 + 3)6 + C. (Substitute u = 2x2 + 3, so du = 4x dx. The integral becomes 14

∫u5 du.)

(b) 13ex3+2 + C. (Substitute u = ex3+2.) (c) 1

4

(ln(x + 2)

)2 + C. (Substitute u = ln(x + 2).)

(d) 25 (1 + x)5/2 − 2

3 (1 + x)3/2 + C. (Substitute u = √1 + x.)

(e)−1

2(1 + x2)+ 1

4(1 + x2)2+ C. (Substitute u = 1 + x2, or u = (1 + x2)−1.)

(f) 215 (4 − x3)5/2 − 8

9 (4 − x3)3/2 + C. (Substitute u = √4 − x3.)

4.∫ x

3

2t − 2

t2 − 2tdt =

∣∣∣∣x3

ln(t2 − 2t) = ln(x2 − 2x) − ln 3 = ln 13 (x2 − 2x), so the given equation reduces to

13 (x2 − 2x) = 2

3x − 1. Hence, x2 − 4x + 3 = 0, with solutions x = 1 and x = 3. But only x = 3 is inthe specified domain. So the solution is x = 3.

6. (a) 1/70 (The integrand can be written as −x4(x5 − 1)13. Then put u = x5 − 1.)(b) 2

√x ln x − 4

√x + C. (Let u = √

x. Then u2 = x, 2udu = dx, etc. Alternatively, use integration byparts.) (c) 8/3. (Let u = √

1 + √x.)

9.7

2. (a)∫ +∞

−∞f (x) dx =

∫ b

a

1

b − adx = 1

b − a

∣∣∣ba

x = 1

b − a(b − a) = 1

(b)∫ +∞

−∞xf (x) dx =

∫ b

a

x

b − adx = 1

2(b − a)

∣∣∣ba

x2 = 1

2(b − a)(b2 − a2) = 1

2(a + b)

(c)1

3(b − a)

b

a

x3 = 1

3

b3 − a3

b − a= 1

3(a2 + ab + b2)

4. The first integral diverges because∫ b

0[x/(1+x2)] dx =

b

0

12 ln(1+x2) = 1

2 ln(1+b2) → ∞ as b → ∞.

On the other hand,∫ b

−b

[x/(1 + x2)] dx =b

−b

1

2ln(1 + x2) = 0 for all b, so the limit as b → ∞ is 0.

6.1

1 + x2≤ 1

x2for x ≥ 1, and

∫ b

1

1

x2dx =

∣∣∣b1

(−1/x) = 1 − 1/b → 1 as b → ∞, so by Theorem 9.7.1

the given integral converges.

8. (a) z =∫ τ

0

1

τe−rs ds = 1

rτ(1 − e−rτ ) (b) z =

∫ τ

0

2(τ − s)

τ 2e−rsds = 2

rτ

(1 − 1

rτ(1 − e−rτ )

)10. Using the answer to Problem 9.6.6(b),

∫ 1

h

ln x√x

dx =∣∣1h

(2√

x ln x −4√

x) = −4−(2√

h ln h−4√

h) →−4 as h → 0+, so the given integral converges to −4. (

√h ln h = ln h/h−1/2 → 0, by l’Hôpital’s rule.)



12. (a) The suggested substitution gives∫ +∞

−∞f (x) dx = 1√

π

∫ +∞

−∞e−u2

du = 1, by (8).

(b) Here∫ +∞

−∞xf (x) dx = 1√

π

∫ +∞

−∞(µ + √

2σu)e−u2du = µ, using part (a) and Example 3.

(c) Here I =∫ +∞

−∞x2f (x) dx = 1√

π

∫ +∞

−∞(2σ 2u2 + 2

√2σµu + µ2)e−u2

du

= 2σ 2

√π

∫ +∞

−∞u2e−u2

du + 2√

2σµ√π

∫ +∞

−∞ue−u2

du + µ2

√π

∫ +∞

−∞e−u2

du = σ 2 + 0 + µ2. (Note how

integration by parts gives∫

u2e−u2du = −1

2ue−u2 +

∫1

2e−u2

du, so∫ +∞

−∞u2e−u2

du = 12

√π .)

9.8

2. (a) K(t) = (K0 − I/δ)e−δt + I/δ (b) (i) K(t) = 200 − 50e−0.05t and K(t) tends to 200 from belowas t → ∞. (ii) K(t) = 200 + 50e−0.05t , and K(t) tends to 200 from above as t → ∞.

4. N(t) = 0.02N(t) + 4 · 104. The solution with N(0) = 2 · 106 is N(t) = 2 · 106(2e0.02t − 1).

6. The percentage surviving after t seconds satisfies p(t) = 100e−δt , where p(7) = 70.5 and so δ =− ln 0.705/7 ≈ 0.05. Thus p(30) = 100e−30δ ≈ 22.3% are still alive after 30 seconds. Because100e−δt = 5 when t ≈ ln 20/0.05 ≈ 60, it takes about 60 seconds to kill 95%.

8. (a) In 1950 there were about 276 thousand. In the next 10 years the number increased by 155 thousand.(b) y → 479.36 as t → ∞. See Fig. M9.8.8 for the graph.

100

200

300

400

500

t5 10 15 20

(1950) (1960) (1970)

Tractors (in 1000)

Figure M9.8.8

10. At about 11:26. (Measuring time in hours, with t = 0 being 12 noon, one has T = k(20−T ) with T (0) =35 and T (1) = 32. So the body temperature at time t is T (t) = 20+15e−kt with k = ln(5/4). Assumingthat the temperature was the normal 37 degrees at the time of death t∗, then t∗ = − ln(17/15)/ ln(5/4) ≈−0.56 hours, or about 34 minutes before 12:00.)


2. (a) e2x + C (b) 12x2 − 25

2 e25 + C (c) − 1

3e−3x + 13e3x + C (d) 2 ln |x + 5| + C



4. (a) 5/4, according to Example 9.7.2. (b)∣∣∣10

120 (1 + x4)5 = 31/20

(c)∣∣∣∞0

5te−t − ∫∞0 5e−t dt = 5

∣∣∣∞0

e−t = −5

(d)∫ e

1 (ln x)2 dx =∣∣∣e1x(ln x)2 − 2

∫ e

1 ln x dx = e − 2∣∣∣e1(x ln x − x) = e − 2

(e)∣∣∣20

29 (x3 + 1)3/2 = 2

9 (93/2 − 1) = 52/9 (f)∣∣∣0−∞

13 ln(e3z + 5) = 1

3 (ln 6 − ln 5) = 13 ln(6/5)

6. F ′(x) = 4(√

x − 1). (Hint:∫ x

4(u1/2 + xu−1/2) du =

x

4

23u3/2 + 2xu1/2 = 8

3x3/2 − 163 − 4x.)

8. C(x) = α

β(eβx − 1) + γ x + C0

10. P ∗ = Q∗ = 5, CS = 50 ln 2 − 25, PS = 1.25

12. (a) x = Ae−3t (b) x = Ae−4t + 3 (c) x = −3

12 − Ae−3t(d) x = Ae− 1

5 t (e) x = Ae−2t + 5/3

(f) x = − 12

1 − Ae− 12 t

= 1

Be− 12 t − 2

(B = 2A)

14. (a) Y = α(a − 1)Y + α(b + I ) (b) Y =(

Y0 − b + I

1 − a

)e−α(1−a)t + b + I

1 − a


40 C H A P T E R 1 0 I N T E R E S T R A T E S A N D P R E S E N T V A L U E S

Chapter 10 Interest Rates and Present Values

This chapter essentially deals with the calculation of interest and present values. Section 10.1 considersdifferent interest periods, and defines the effective yearly interest rate. Section 10.2 considers the case wheninterest is compounded so frequently that one may as well treat time as continuous and have “continuouscompounding” of interest.

The idea of present values is important for economists and it is discussed in Section 10.3.Section 10.4 is concerned with geometric series and their sums. The condition for an infinite geometric

series to have a convergent sum is presented. Of course, this topic is not specific to finance, but we includeit here because it can be motivated by the need to discount payment streams extending into the indefinitefuture. When the rate of discount is fixed, this is treated in Section 10.5, where the present and future valuesof annuities are also derived.

Mortgage repayments are the topic in Section 10.6, which also includes a short subsection on depositswithin an interest period. The final Section 10.7 gives a brief discussion of internal rates of return.


10.1

2. (a) 5000(1+0.03)10 ≈ 6719.58 (b) 37.17 years. (5000(1.03)t = 3·5000, so t = ln 3/ ln 1.03 ≈ 37.17.)

4. (a) (i) After 2 years: 2000(1.07)2 = 2289.80 (ii) After 10 years: 2000(1.07)10 ≈ 3934.30(b) t = ln 3/ ln 1.07 ≈ 16.2 years.

6. The effective yearly rate for alternative (ii) is (1 + 0.2/4)4 − 1 = 1.054 − 1 ≈ 0.2155 > 0.215, soalternative (i) is (slightly) cheaper.

8. Let the nominal yearly rate be r . By (2), 0.28 = (1 + r/4)4 − 1, so r = 4(4√

1.28 − 1) ≈ 0.25, or 25%.

10.2

2. (a) (i) 1000(1 + 0.05)10 ≈ 1629 (ii) 1000(1 + 0.05/12)120 ≈ 1647 (iii) 1000e0.05·10 ≈ 1649(b) (i) 1000(1 + 0.05)50 ≈ 11467 (ii) 1000(1 + 0.05/12)600 ≈ 12119 (iii) 1000e0.05·50 ≈ 12182

4. e−0.1t∗ = 1/10, so −0.1t∗ = − ln 10, hence t∗ = ln 10/0.1 ≈ 23.

6. h′(u) = u/(1 + u)2 > 0 for u > 0, so h(u) > 0 for u > 0, implying that g′(x)/g(x) = h(r/x) > 0for all x > 0. So g(x) is strictly increasing for x > 0. Because g(x) → er as x → ∞, it follows thatg(x) < er for all x > 0. Continuous compounding of interest is best for the lender.

10.3

2. (i) The present value is 50 000 · 1.0575−5 ≈ 37 806.64. (ii) 50 000 · e−0.0575·5 ≈ 37 506.83

10.4

2. (a) (1/5)/(1 − 1/5) = 1/4 (b) 0.1/(1 − 0.1) = 1/9 (c) 517/[1 − (1.1)−1] = 5687(d) a/[1 − (1 + a)−1] = 1 + a (e) 5/(1 − 3/7) = 35/4


C H A P T E R 1 0 I N T E R E S T R A T E S A N D P R E S E N T V A L U E S 41

4. (a) Quotient k = 1/p. Converges to 1/(p − 1) for |p| > 1.(b) Quotient k = 1/

√x. Converges to x

√x/(

√x − 1) for

√x > 1, that is, for x > 1.

(c) Quotient k = x2. Converges to x2/(1 − x2) for |x| < 1.

6. Let x denote the number of years beyond 1971 that the Earth’s extractable resources of iron will last. Thenwe get 794+794·1.05+· · ·+794·(1.05)x = 249·103. Using (3), 794[1−(1.05)x+1]/(1−1.05) = 249·103

or (1.05)x+1 = 249 · 103 · 0.05/794 ≈ 16.68. Using a calculator, we find x ≈ (ln 16.68/ ln 1.05) − 1 ≈56.68, so the resources will be exhausted part way through the year 2028.

8. (a) f (t) = P(t)e−rt

1 − e−rt= P(t)

ert − 1(b) Here f ′(t) = P ′(t)(ert − 1) − P(t)rert

(ert − 1)2, and t∗ > 0 can only

maximize f (t) if f ′(t∗) = 0, that is, if P ′(t∗)(ert∗ − 1) = rP (t∗)ert∗ , which implies that P ′(t∗) =r P (t∗)/(1 − e−rt∗). (c) Using l’Hôpital’s rule, P ′(t∗)/P (t∗) → 1/t∗ as r → 0.

10.5

2. (a) 10 years ago the amount was: 100 000(1.04)−10 ≈ 67556.42

(b) 10000(1.063 + 1.062 + 1.06 + 1) = 10 0001.064 − 1

1.06 − 1≈ 43746.16

4. Offer (a) is better. The second offer has present value 46001 − (1.06)−5

1 − (1.06)−1≈ 20 540.

6. If the largest amount is a, then according to formula (4), a/r = K , so that a = rK .

8. PDV = ∫ 150 500e−0.06t dt = 500

∣∣150

−10.06e−0.06t = 500

0.06

[1 − e−0.9

] ≈ 4945.25.

FDV = e0.06·15PDV = e0.9PDV ≈ 2.4596 · 4945.25 ≈ 12163.3.

10.6

2. Using (2) we get a = (0.07/12) · 80000

1 − (1 + 0.07/12)−120≈ 928.87.

4. Schedule (b) has present value12 000 · 1.115

0.115[1 − (1.115)−8] ≈ 67 644.42.

Schedule (c) has present value 22 000 + 7000

0.115[1 − (1.115)−12] ≈ 66 384.08.

Thus schedule (c) is cheapest. When the interest rate becomes 12.5 %, schedules (b) and (c) have presentvalues equal to 65907.61 and 64374.33, respectively.

10.7

2. Equation (1) is herea

1 + r+ a

(1 + r)2+ · · · = −a0, which yields a/r = −a0, so r = −a/a0.

4. $ 1.55 million. (400 000(1/1.175 + (1/1.175)2 + · · · + (1/1.175)7) ≈ 1 546 522.94.)

6. Applying (10.5.2) with a = 1000 and n = 5 gives the equation P5 = (1000/r)[1 − 1/(1 + r)5

] = 4340to be solved for r . For r = 0.05%, the present value is $4329.48; for r = 0.045%, the present value is$4389.98. Because dP5/dr < 0, it follows that p is a little less than 5%.


42 C H A P T E R 1 0 I N T E R E S T R A T E S A N D P R E S E N T V A L U E S


2. (a) 8000 · 1.053 = 9261 (b) 8000 · 1.0513 ≈ 15085.19 (c) (1.05)t∗ = 4, so t∗ = ln 4/1.05 ≈ 28.5

4. 15 000e0.07·12 ≈ 34745.50

6. (a) 44/(1 − 0.56) = 100 (b) 20/[1 − (1.2)−1] = 120 (c) 3/(1 − 2/5) = 5(d) 400/(1 − 20−1) = 8000/19

8. (a) 5000(1.04)4 = 5849.29. (b)5000

0.04[(1.04)4 − 1] = 21 232.32

(c) The last payment will be on 1st January 2006, when the initial balance of 10 000 will have earnedinterest for 10 years. So K must solve 10 000 · (1.04)10 + K[(1.04)8 − 1]/0.04 = 70 000. We find thatK ≈ 5990.49.

10. (a) Present value:3200

0.08[1 − (1.08)−10] = 21 472.26.

(b) Present value: 7000 + 3000

0.08[1 − 1.08−5] = 18 978.13.

(c) Four years ahead the present value is4000

0.08[1 − (1.08)−10] = 26 840.33. The present value when

Lucy makes her choice is 26 840.33 · 1.08−4 = 19 728.44. She should choose option (a) .


C H A P T E R 1 1 F U N C T I O N S O F M A N Y V A R I A B L E S 43

Chapter 11 Functions of Many VariablesIn principle, this chapter represents the first occasion on which the student will ever have seen functions ofmany variables. For this reason, the rather slow start provided by Section 11.1 does seem required. In fact,the first three sections deal exclusively with functions of two variables. Though definition (1) is abstract, it isimmediately followed by concrete examples. As usual, first comes the mechanical Example 1, which preparesthe student for problems 1–4. Then come, in rapid succession, a demand function with both price and incomeas variables, a Cobb–Douglas production function, and an example that prepares for the later discussion ofhomogeneous functions. Problem 5 is especially relevant at this stage. The final discussion of domains mayseem to deserve lower priority, but difficulties may arise later if this topic is neglected.

Section 11.2 introduces the key idea of partial differentiation. We try to make it as easy as possibleby borrowing from earlier work for functions of one variable. In addition to mastering the technique forpartially differentiating simple functions, the student should be told how important it is to understand thebasic definitions of the partial derivatives given in (2) and (3), as well as the approximations in (4) to (6). Inparticular, economists often interpret the partial derivative f ′

1(x, y) as approximately equal to the change inf (x, y) that results from increasing x by one unit while holding y constant. (Of course, it is better in principleto interpret f ′

1(x, y)h as approximately equal to f (x + h) − f (x, y), for h small.)After the mechanics of partial derivatives, Section 11.3 turns to geometric representations of functions.

Graphs are useful for functions of one variable. They are less useful for functions of two variables (or more)because of the difficulty of drawing in three (or more) dimensions. Nevertheless, for many functions of twovariables, level curves are a useful device that often arises in economic contexts—especially as indifferencecurves, isoquants, etc. The relationship between partial derivatives, graphs, and level curves is important.

Section 11.4 explains how an equation in three variables can be represented, in principle, by a surface inthree dimensional space. An easy example is the budget plane if there are three commodities. The distanceformula in three dimensional space is derived and the equation for a sphere in three dimensions follows easily.

Following all the work in Sections 11.1–11.3 with functions of two variables, we hope that the extensionto n variables in Sections 11.5 and 11.6 will seem routine. The general Cobb–Douglas function and its log-linear transformation are important in economics. Example 11.5.2 concerns the arithmetic, geometric, andharmonic means. Our discussion of continuity and the Euclidean n-dimensional space �n (for n > 3) is brief.These concepts will tax the powers of most undergraduate students if taken beyond this very intuitive level.

The only really new concept in Section 11.6 is the Hessian matrix. Young’s Theorem 11.6.1 is statedcarefully, as is the formal definition of the partial derivative, and the concept of a Ck function. The latter isused occasionally in later chapters.

Section 11.7 brings in economic applications, starting with marginal products. The notation F ′K , F ′

L, F ′T

is preferred for these (when K is capital, L is labour, and T is land) because it makes clear what factor is beingvaried in each case. Brief mention is made of complementary factors and, more importantly, of diminishingmarginal products. These are applications of second partial derivatives, of course.

This chapter ends with a brief discussion of partial elasticities in Section 11.8. These, of course, arise inmany economic contexts.


11.12. f (0, 1) = 0, f (−1, 2) = −4, f (104, 10−2) = 1, f (a, a) = a3, f (a + h, b) = (a + h)b2 = ab2 + hb2,

and f (a, b + k) − f (a, b) = 2abk + ak2.


44 C H A P T E R 1 1 F U N C T I O N S O F M A N Y V A R I A B L E S

4. (a) f (−1, 2) = 1, f (a, a) = 4a2, f (a + h, b) − f (a, b) = 2(a + b)h + h2

(b) f (tx, ty) = (tx)2 + 2(tx)(ty) + (ty)2 = t2(x2 + 2xy + y2) = t2f (x, y) for all t , including t = 2.

6. (a) y �= x − 2 (b) x2 + y2 ≤ 2 (c) 1 ≤ x2 + y2 ≤ 4. The domains in (b) and (c) are the shaded setsshown in Figs. M11.1.6b and c.

y

x

x2 + y2 ≤ (√2)2

y

x1 2x

y

z

Figure M11.1.6b Figure M11.1.6c Figure M11.3.6

11.2

2. (a) ∂z/∂x = 2x, ∂z/∂y = 6y (b) ∂z/∂x = y, ∂z/∂y = x

(c) ∂z/∂x = 20x3y2 − 2y5, ∂z/∂y = 10x4y − 10xy4 (d) ∂z/∂x = ∂z/∂y = ex+y

(e) ∂z/∂x = yexy , ∂z/∂y = xexy (f) ∂z/∂x = ex/y, ∂z/∂y = −ex/y2

(g) ∂z/∂x = ∂z/∂y = 1/(x + y) (h) ∂z/∂x = 1/x, ∂z/∂y = 1/y

4. (a) z′x = 3, z′

y = 4, and z′′xx = z′′

xy = z′′yx = z′′

yy = 0(b) z′

x = 3x2y2, z′y = 2x3y, z′′

xx = 6xy2, z′′yy = 2x3, and z′′

xy = z′′yx = 6x2y

(c) z′x = 5x4 − 6xy, z′

y = −3x2 + 6y5, z′′xx = 20x3 − 6y, z′′

yy = 30y4, and z′′xy = z′′

yx = −6x

(d) z′x = 1/y, z′

y = −x/y2, z′′xx = 0, z′′

yy = 2x/y3, and z′′xy = z′′

yx = −1/y2

(e) z′x = 2y(x + y)−2, z′

y = −2x(x + y)−2, z′′xx = −4y(x + y)−3, z′′

yy = 4x(x + y)−3, and z′′xy =

z′′yx = 2(x − y)(x + y)−3 (f) z′

x = x(x2 + y2)−1/2, z′y = y(x2 + y2)−1/2, z′′

xx = y2(x2 + y2)−3/2,z′′yy = x2(x2 + y2)−3/2, and z′′

xy = z′′yx = −xy(x2 + y2)−3/2

6. (a) F ′S = 2.26·0.44S−0.56E0.48 = 0.9944S−0.56E0.48, F ′

E = 2.26·0.48S0.44E−0.52 = 1.0848S0.44E−0.52

(b) SF ′S + EF ′

E = S · 2.26 · 0.44S−0.56E0.48 + E · 2.26 · 0.48S0.44E−0.52 = 0.44 F + 0.48 F = 0.92 F ,so k = 0.92.

8. Here ∂z/∂x = x/(x2 + y2), ∂z/∂y = y/(x2 + y2), ∂2z/∂x2 = (y2 − x2)/(x2 + y2)2, and ∂2z/∂y2 =(x2 − y2)/(x2 + y2)2. Thus, ∂2z/∂x2 + ∂2z/∂y2 = 0.

11.3

2. (a) A straight line through (0, 2, 3) parallel to the x-axis.(b) A plane parallel to the z-axis whose intersection with the xy-plane is the line y = x.

4. f (x, y) = ex2−y2 + (x2 − y2)2 = ec + c2 when x2 − y2 = c, so the last equation represents a level curveof f at height ec + c2.


C H A P T E R 1 1 F U N C T I O N S O F M A N Y V A R I A B L E S 45

6. Generally, the graph of g(x, y) = f (x) in 3-space consists of a surface traced out by moving the graphof z = f (x) parallel to the y-axis in both directions. The graph of g(x, y) = x is the plane through they-axis at a 45◦ angle with the xy-plane. The graph of g(x, y) = −x3 is shown in Fig. M11.3.6. (Only aportion of the the unbounded graph is indicated, of course.)

8. (a) f (2, 3) = 8. f (x, 3) = 8 has the solutions x = 2 and x = 5 (b) As y varies with x = 2 fixed,the minimum of f (2, y) is 8 when y = 3. (c) A: f ′

1(x, y) > 0, f ′2(x, y) > 0. B: f ′

1(x, y) < 0,f ′

2(x, y) < 0. C: f ′1(x, y) = 0, f ′

2(x, y) = 0. At A, f ′1 ≈ 2/1 = 2 and f ′

2 ≈ 2/0.6 = 10/3.

10. F(1, 0) = F(0, 0) + ∫ 10 F ′

1(x, 0) dx ≥ 2, F(2, 0) = F(1, 0) + ∫ 21 F ′

1(x, 0) dx ≥ F(1, 0) + 2,

F(0, 1) = F(0, 0) + ∫ 10 F ′

2(0, y) dy ≤ 1, F(1, 1) = F(0, 1) + ∫ 10 F ′

1(x, 1) dx ≥ F(0, 1) + 2,

F(1, 1) = F(1, 0) + ∫ 10 F ′

2(1, y) dy ≤ F(1, 0) + 1.

11.4

2. d = √(4 − (−1))2 + (−2 − 2)2 + (0 − 3)2 = √

25 + 16 + 9 = √50 = 5

√2.

4. (x − 2)2 + (y − 1)2 + (z − 1)2 = 25

6. (x − 4)2 + (y − 4)2 + (z − 12 )2 measures the square of the distance from the point (4, 4, 1

2 ) to the point(x, y, z) on the paraboloid.

11.5

2. (a) F(K + 1, L, T ) − F(K, L, T ) is the increase in output from increasing capital input by one unit.(b) F(K, L, T ) = AKaLbT c, where A, a, b, and c are constants, A > 0.(c) F(tK, tL, tT ) = ta+b+cF (K, L, T )

4. You drive (5/60) ·0+(10/60) ·30+(20/60) ·60+(15/60) ·80 = 45 kilometres in 5+10+20+15 = 50minutes, so the average speed is 45 × 60/50 = 54 kph.

6. (a) Each machine would produce 60 units per day, so each unit produced would require 480/60 = 8minutes. (b) Total output is

∑ni=1(T /ti) = T

∑ni=1(1/ti). If all n machines were equally efficient,

the time needed for each unit would be nT/T∑n

i=1(1/ti) = n/∑n

i=1(1/ti), the harmonic mean oft1, . . . , tn.

11.6

2. (a) f ′1 = 2x, f ′

2 = 3y2, and f ′3 = 4z3 (b) f ′

1 = 10x, f ′2 = −9y2, and f ′

3 = 12z3

(c) f ′1 = yz, f ′

2 = xz, and f ′3 = xy (d) f ′

1 = 4x3/yz, f ′2 = −x4/y2z, and f ′

3 = −x4/yz2

(e) f ′1 = 12x(x2 + y3 + z4)5, f ′

2 = 18y2(x2 + y3 + z4)5, and f ′3 = 24z3(x2 + y3 + z4)5

(f) f ′1 = yzexyz, f ′

2 = xzexyz, and f ′3 = xyexyz

4. First-order partials: w′1 = 3yz + 2xy − z3, w′

2 = 3xz + x2, w′3 = 3xy − 3xz2. Second-order partials:

w′′11 = 2y, w′′

12 = w′′21 = 3z + 2x, w′′

13 = w′′31 = 3y − 3z2, w′′

22 = 0, w′′23 = w′′

32 = 3x, w′′33 = −6xz.

6. (a)

⎛⎝ 2a 0 00 2b 00 0 2c

⎞⎠ (b)

⎛⎝ a(a − 1)g/x2 abg/xy acg/xz

abg/xy b(b − 1)g/y2 bcg/yz

acg/xz bcg/yz c(c − 1)g/z2

⎞⎠, in concise form.

8. f ′x = yzxyz−1, f ′

y = zyz−1(ln x)xyz

, f ′z = yz(ln x)(ln y)xyz


46 C H A P T E R 1 1 F U N C T I O N S O F M A N Y V A R I A B L E S

11.7

2. Simplified answers: (a) KY ′K + LY ′

L = aY (b) KY ′K + LY ′

L = (a + b)Y (c) KY ′K + LY ′

L = Y

4. ∂D/∂p and ∂E/∂q are normally negative, because the demand for a commodity goes down when theprice of that commodity increases. If the commodities are substitutes, this means that demand increaseswhen the price of the other good increases. So the usual signs are ∂D/∂q > 0 and ∂E/∂p > 0.

6. KY ′K + LY ′

L = mY . (Y ′K = −(m/ρ)a(−ρ)AeλtK−ρ−1[aK−ρ + (1 − a)L−ρ]−(m/ρ)−1 and

Y ′L = −(m/ρ)(1 − a)(−ρ)AeλtL−ρ−1[aK−ρ + (1 − a)L−ρ]−(m/ρ)−1, so

KY ′K + LY ′

L = −(m/ρ)(−ρ)[aK−ρ + (1 − a)L−ρ]−m/ρ = mY.

11.8

2. Let z = ug with u = axd1 + bxd

2 + cxd3 . Then El1z = Elu ugEl1u = g(x1/u)adxd−1

1 = adgxd1 /u.

Similarly, El2z = bdgxd2 /u and El3z = cdgxd

3 /u, so El1z+El2z+El3z = dg(axd1 +bxd

2 +cxd3 )/u = dg.

(This result follows easily from the fact that the function is homogeneous of degree dg (see Problem12.7.2(b) and the elasticity form (12.7.3) of the Euler equation.)

4.∂

∂m

(pD(p, m)

m

)=p

mD′m(p, m) − D(p, m)

m2= pD(p, m)

m2[Elm D(p, m)−1] > 0 iff Elm D(p, m) > 1,

so pD/m increases with m if Elm D > 1. (Using the formulas in Problem 7.7.10, the result also followsfrom the fact that Elm(pD(p, m)/m) = Elm D(p, m) − Elm m = Elm D(p, m) − 1.)


2. f (−1, 2) = −10, f (2a, 2a) = −4a2, f (a, b+ k)−f (a, b) = −6bk −3k2, f (tx, ty)− t2f (x, y) = 0.

4. (a) ∂Y/∂K ≈ 0.083K0.356S0.562 and ∂Y/∂S ≈ 0.035K1.356S−0.438.(b) The catch becomes 21.356+0.562 = 21.918 ≈ 3.779 times higher.

6. F ′K = aF/K , F ′

L = bF/L, and F ′M = cF/M , so KF ′

K + LF ′L + MF ′

M = (a + b + c)F .

8. (a) g(2, 1, 1) = −2, g(3, −4, 2) = 352, and g(1, 1, a + h) − g(1, 1, a) = 2ah + h2 − h.(b) ∂g/∂x = 4x − 4y − 4, ∂g/∂y = −4x + 20y − 28, ∂g/∂z = 2z − 1. The Hessian matrix of the

second-order partials is:

⎛⎝ 4 −4 0−4 20 0

0 0 2

⎞⎠.

10. (a) ∂z/∂x = 10xy4(x2y4 + 2)4 (b)√

K(∂F/∂K) = 2√

K(√

K + √L)(1/2

√K) = √

K + √L

(c) KF ′K + LF ′

L = K(1/a)aKa−1(Ka + La)1/a−1 + L(1/a)aLa−1(Ka + La)1/a−1 =(Ka + La)(Ka + La)1/a−1 = F (d) ∂g/∂t = 3/w + 2wt , so ∂2g/∂w∂t = −3/w2 + 2t

(e) g′3 = t3(t

21 + t2

2 + t23 )−1/2 (f) f ′

1 = 4xyz + 2xz2, f ′′13 = 4xy + 4xz

12. If x − y = c, then F(x, y) = ln(x − y)2 + e2(x−y) = ln c2 + e2c, a constant.

14. (a) Elx z = 3, Ely z = −4 (b) Elx z = 2x2

(x2 + y2) ln(x2 + y2), Ely z = 2y2

(x2 + y2) ln(x2 + y2)

(c) Elx z = Elx(exey) = Elx ex = x, Ely z = y (d) Elx z = x2/(x2 + y2), Ely z = y2/(x2 + y2)


C H A P T E R 1 2 T O O L S F O R C O M P A R A T I V E S T A T I C S 47

Chapter 12 Tools for Comparative Statics

Chapter 12 gathers together a number of tools and techniques from the calculus of many variables that are muchused by economists. It also introduces important concepts such as elasticity of substitution, homogeneousand homothetic functions, linear approximations, and differentials.

First, however, Section 12.1 deals with the chain rule for differentiating functions of many variables,when these variables all depend on a single variable such as time. Of course, it begins with the two variablecase, and Examples 1 and 2 where the validity of the rule is checked by direct differentiation. Examples 3and 4 are just two of the many obvious economic applications that spring to mind immediately. Finally, thesection closes with an attempt (in small print) to justify the chain rule.

Next, Section 12.2 considers chain rules for functions of many variables that are themselves functions ofmany variables. It is necessary to learn how to keep notation straight, and under control!

Section 12.3 deals with implicit differentiation, generalizing the results from Section 7.1. It introducesformula (1), giving the slope of a level curve for the two-variable case. This formula is then applied mech-anically in Examples 1 to 3, and to the economic Example 4 on the incidence of taxation in a simple modelof supply and demand. Next comes a formula for the second derivative. For those students who have alreadylearned about determinants, the second derivative can also be expressed as a bordered Hessian (though we donot use this term).

Section 12.4 develops more general results on implicit differentiation. Two interesting examples are astandard profit maximization model, and then a simple search model.

Section 12.5 introduces the elasticity of substitution, with standard applications to the Cobb–Douglasand CES (constant elasticity of substitution) functions.

Homogeneous functions receive much attention in economics. Section 12.6 introduces them for the caseof two variables, along with Euler’s theorem and other properties. Some geometric properties that may behelpful are discussed.

Section 12.7 considers homogeneous functions of many variables. After proving that a function is homo-geneous if and only if it satisfies the Euler equation, we show that the other relevant properties of functions oftwo variables generalize appropriately. Economic applications are then considered. In particular, the relation-ship between homogeneity and returns to scale is considered, as well as homogeneity of demand functions.The section concludes with an introduction to homothetic functions, and how they generalize homogeneousfunctions.

Next, Section 12.8 considers linear approximations and tangent planes.

Section 12.9 turns to differentials of functions of several variables. Rules for differentials are presented,as well as an explanation of how they are natural counterparts of rules for derivatives. The invariance of thedifferential, however, deserves more careful consideration, though most economists have neglected it.

Section 12.10, on systems of equations, begins with an elementary discussion of degrees of freedom,and the associated counting rule. Some examples illustrate difficulties with the counting rule for nonlinearfunctions.1

1 We do not discuss the concept of functional dependence. Most texts in mathematics for economists state erro-neously that functional dependence of n functions is equivalent to the vanishing of the associated Jacobian determinant.For a correct account, see J. E. Marsden and M. J. Hoffman: Elementary Classical Analysis, 2. ed. W. H. Freeman andCo. (1993).


48 C H A P T E R 1 2 T O O L S F O R C O M P A R A T I V E S T A T I C S

The main part of Section 12.11 consists of Examples 1–3 showing how to find partial derivatives fromdifferentials. Whereas Examples 1 and 2 are purely mechanical, with no economic distractions, Example 3 isbased on a rather standard macroeconomic model.

The general case of a system of structural equations is briefly discussed, as is the distinction betweenendogenous and exogenous variables. Finally, the concept of reduced form is defined in this setting.


12.1

2. (a)dz

dt=(

ln y + y

x

)· 1 +

(x

y+ ln x

)1

t= ln(ln t) + ln t

t + 1+ t + 1

t ln t+ ln(t + 1)

t

(b)dz

dt= 1

xAaeat + 1

yBbebt = a + b

4. dY/dt = (10L − 1

2K−1/2)

0.2 + (10K − 1

2L−1/2)

0.5e0.1t = 35 − 7√

5/100 when t = 0 and soK = L = 5.

6. Let U(x) = u(x, h(x)) = ln[xα + (ax4 + b)α/3] − α3 ln(ax4 + b).

Then U ′(x) = αxα−1(3b − ax4)

3[xα + (ax4 + b)α/3](ax4 + b). So U ′(x∗) = 0 at x∗ = 4

√3b/a, whereas U ′(x) > 0 for

x < x∗ and U ′(x) < 0 for x > x∗. Hence x∗ maximizes U .

12.22. (a) ∂z/∂t = y2 + 2xy2ts = 5t4s2 + 4t3s4, ∂z/∂s = y22s + 2xyt2 = 2t5s + 4t4s3

(b)∂z

∂t= 2(1 − s)ets+t+s

(et+s + ets)2and

∂z

∂s= 2(1 − t)ets+t+s

(et+s + ets)2

4. ∂z/∂t1 = F ′(x)f ′1(t1, t2), ∂z/∂t2 = F ′(x)f ′

2(t1, t2)

6.∂C

∂p1= a

∂Q1

∂p1+ b

∂Q2

∂p1+ 2cQ1

∂Q1

∂p1= −α1A(a + 2cAp

−α11 p

β12 )p

−α1−11 p

β12 + α2bBp

α2−11 p

−β22

∂C

∂p2= β1A(a + 2cAp

−α11 p

β12 )p

−α11 p

β1−12 − β2bBp

α21 p

−β2−12

8. (a)∂u

∂r= ∂f

∂x

∂x

∂r+ ∂f

∂y

∂y

∂r+ ∂f

∂z

∂z

∂r+ ∂f

∂w

∂w

∂r(b)

∂u

∂r= yzw + xzw + xyws + xyz(1/s) = 28

12.32. See the answers to Problem 7.1.1. (For (a): Put F(x, y) = x2y. Then F ′

1 = 2xy, F ′2 = x2, F ′′

11 =2y, F ′′

12 = 2x, F ′′22 = 0, so y ′ = −F ′

1/F′2 = −2xy/x2 = −2y/x. Moreover, according to (3),

y ′′ = −(1/(F ′2)

3)[F ′′

11(F′2)

2 − 2F ′′12F

′1F

′2 + F ′′

22(F′1)

2] = −(1/x6)[2yx4 − 2(2x)(2xy)x2] = 6y/x2.)

4. h′(x) = − 6x − 3y2

−6xy + 3y2 + 6y= −1 at (x, y) = (1, 1).

6. Differentiating the equation w.r.t. x gives (i) 1 − az′x = f ′(y − bz)(−bz′

x). Differentiating w.r.t. y

gives (ii) −az′y = f ′(y − bz)(1 − bz′

y). If bz′x �= 0, solving (i) for f ′ and inserting it into (ii) yields

az′x + bz′

y = 1. If bz′x = 0, then (i) implies az′

x = 1. But then z′x �= 0, so b = 0 and then again

az′x + bz′

y = 1.



12.4

2. Differentiating partially w.r.t. x yields (∗) 3x2 + 3z2z′x − 3z′

x = 0, so z′x = x2/(1 − z2). By symmetry,

z′y = y2/(1 − z2). To find z′′

xy , differentiate (∗) w.r.t. y to obtain 6zz′yz

′x + 3z2z′′

xy − 3z′′xy = 0, so

z′′xy = 2zx2y2/(1 − z2)3. (Alternatively, differentiate z′

x = x2/(1 − z2) w.r.t. y.)

4. z′x = −yxy−1 + zx ln z

yz ln y + xzx−1and z′

y = −xy ln x + zyz−1

yz ln y + xzx−1

6. (a) F(1, 3) = 4. The equation for the tangent is y − 3 = −F ′x(1, 3)

F ′y(1, 3)

(x − 1) with F ′x(1, 3) = 10 and

F ′y(1, 3) = 5, so y = −2x + 5. (b)

∂y

∂K= αy

K(1 + 2c ln y),

∂y

∂L= βy

L(1 + 2c ln y)

12.5

2. (a) Ryx = (x/y)a−1 = (y/x)1−a (b) σyx = ElRyx(y/x) = ElRyx

(Ryx)1/(1−a) = 1/(1 − a)

12.6

2. x(tp, tr) = A(tp)−1.5(tr)2.08 = At−1.5p−1.5 t2.08 r2.08 = t−1.5 t2.08Ap−1.5r2.08 = t0.58x(p, r), so thefunction is homogeneous of degree 0.58 . (Alternatively, use the result in Example 11.1.4.)

4. Checking definition (1) shows that f is homogeneous of degree 0. Then using the partial derivativesfound in Example 11.2.1(b), we confirm that xf ′

1(x, y) + yf ′2(x, y) = 0.

6. Definition (1) requires that for some number k one has, t3x3 + t2xy = tk(x3 + xy) for all t > 0 andall (x, y). In particular, for x = y = 1, we must have t3 + t2 = 2tk . For t = 2, we get 12 = 2 · 2k , or2k = 6. For t = 4, we get 80 = 2 · 4k , or 4k = 40. But 2k = 6 implies 4k = 36. So the two values of k

must actually be different, implying that f is not homogeneous of any degree.

8. Let C and D denote the the numerator and the denominator in the expression for σyx in Problem 12.5.3.Because F is homogeneous of degree one, Euler’s theorem implies that C = −F ′

1F′2F , and (∗) implies

that xF ′′11 = −yF ′′

12 and yF ′′22 = −xF ′′

21 = −xF ′′12. Hence, D = xy

[(F ′

2)2F ′′

11−2F ′1F

′2F

′′12+(F ′

1)2F ′′

22

] =−F ′′

12

[y2(F ′

2)2 + 2xyF ′

1F′2 + x2(F ′

1)2] = −F ′′

12(xF ′1 + yF ′

2)2 = −F ′′

12F2, using Euler’s theorem again.

Hence σxy = C/D = (−F ′1F

′2F)/(−F ′′

12F2) = F ′

1F′2/FF ′′

12.

12.7

2. (a) The function is homogeneous of degree 1 because

F(tx1, tx2, tx3) = (tx1tx2tx3)2

(tx1)4 + (tx2)4 + (tx3)4

(1

tx1+ 1

tx2+ 1

tx3

)= t6(x1x2x3)

2

t4(x41 + x4

2 + x43)

1

t

(1

x1+ 1

x2+ 1

x3

)= tF (x1, x2, x3)

(b) G(tx1, tx2, tx3) = [a(tx1)

d + b(tx2)d + c(tx3)

d]g = [

td (axd1 + bxd

2 + cxd3 )]g

= tdgG(x1, x2, x3), so G is homogeneous of degree dg.



4. v′i = u′

i − a/(x1 + · · · + xn), so∑n

i=1 xiv′i = ∑n

i=1 xiu′i −∑n

i=1 axi/(x1 + · · · + xn) =a − [a/(x1 + · · · + xn)]

∑ni=1 xi = a − a = 0. By Euler’s theorem, v is homogeneous of degree 0.

6. Let � = ln C(tw, y) − ln C(w, t). It suffices to prove that � = ln t , because then C(tw, y)/C(w, y) =ln et = t . We find that

� =n∑

i=1

ai[ln(twi) − ln wi] + 1

2

n∑i,j=1

aij [ln(twi) ln(twj ) − ln wi ln wj ] + ln y

n∑i=1

bi[ln(twi) − ln wi]

Because ln(twi) − ln wi = ln t + ln wi − ln wi = ln t and also ln(twi) ln(twj ) − ln wi ln wj = (ln t)2 +ln t ln wi + ln t ln wj , this can be reduced to

� = ln t

n∑i=1

ai + 1

2(ln t)2

n∑i,j=1

aij + 1

2(ln t)

n∑j=1

ln wi

n∑i=1

aij + 1

2(ln t)

n∑i=1

ln wj

n∑j=1

aij + ln y ln t

n∑i=1

bi

Finally, this simplifies to � = ln t + 0 + 0 + 0 + 0 = ln t because of the specified restrictions on theparameters ai , aij , and bi .

12.82. f (x, y) ≈ Axa

0 yb0 + aAxa−1

0 yb0 (x − x0) + bAxa

0 yb−10 (y − y0) = Axa

0 yb0

(1 + a

x − x0

x0+ b

y − y0

y0

)4. f (0.98, −1.01) ≈ −5 − 6(0.98 − 1) + 9(−1.01 + 1) = −4.97. The exact value is −4.970614, so the

error is 0.000614.

6. v(1.01, 0.02) ≈ v(1, 0) + v′1(1, 0) · 0.01 + v′

2(1, 0) · 0.02 = −1 − 1/150

8. g(0) = f (x0), g(1) = f (x). Using formula (12.2.3), it follows thatg′(t) = f ′

1(x0 + t (x − x0))(x1 − x0

1 ) + · · · + f ′n(x

0 + t (x − x0))(xn − x0n). Putting t = 0 gives

g′(0) = f ′1(x

0)(x1 − x01 ) + · · · + f ′

n(x0)(xn − x0

n), and the conclusion follows.

12.9

2. (a) dz = 3x2 dx + 3y2 dy (b) dz = ey2(dx + 2xy dy) (c) dz = 2

x2 − y2(x dx − y dy)

4. Let T (x, y, z) = [x2 + y2 + z2]1/2 = u1/2, where u = x2 + y2 + z2. Then dT = 12u−1/2du =

12u−1/2(2xdx + 2ydy + 2zdz). For x = 2, y = 3, and z = 6, we have u = 49, T = 7 and dT =17 (xdx + ydy + zdz) = 1

7 (2dx + 3dy + 6dz). Thus, T (2 + 0.01, 3 − 0.01, 6 + 0.02) ≈ T (2, 3, 6) +17 [2 · 0.01 + 3(−0.01) + 6 · 0.02] = 7 + 1

7 · 0.11 ≈ 7.015714. (A calculator or computer gives a betterapproximation:

√49.2206 ≈ 7.015739.)

6. (a) dX = AβNβ−1e�tdN + ANβ�e�tdt

(b) dX1 = BEXE−1N1−EdX + B(1 − E)XEN−EdN

8. d(ln z) = a1d(ln x1) + · · · + and(ln xn), so dz/z = a1dx1/x1 + a2dx2/x2 + · · · + andxn/xn.

12.102. There are 6 variables Y , C, I , G, T , and r , and 3 equations. So there are 6 − 3 = 3 degrees of freedom.



12.11

2. (a) Differentiating yields: u3dx + x3u2du + dv = 2y dy and 3v du + 3u dv − dx = 0.Solving for du and dv yields, with D = 9xu3 − 3v,

du = D−1(−3u4 − 1) dx + D−16yu dy and dv = D−1(3xu2 + 3u3v) dx + D−1(−6yv) dy

(b) u′x = D−1(−3u4 − 1), v′

x = D−1(3xu2 + 3u3v) (c) u′x = 283/81 and v′

x = −64/27

4. With a fixed, I ′(r) dr = S ′(Y ) dY and a dY + L′(r) dr = dM . Solving for dY and dr in terms of dM

gives∂Y

∂M= I ′(r)

aI ′(r) + L′(r)S ′(Y )and

∂r

∂M= S ′(Y )

aI ′(r) + L′(r)S ′(Y )

6. (a) Differentiation yields the equations: dY = dC + dI + dG, dC = F ′Y dY + F ′

T dT + F ′r dr , and

dI = f ′Y dY + f ′

r dr . Hence, dY = (F ′

T dT + dG + (F ′r + f ′

r ) dr)/(1 − F ′

Y − f ′Y ).

(b) ∂Y/∂T = F ′T /(1 − F ′

Y − f ′Y ) < 0, so Y decreases as T increases. But if dT = dG with dr = 0,

then dY = (1 + F ′

T

)dT /(1 − F ′

Y − f ′Y ), which is positive provided that F ′

T > −1.

8. (a) There are 3 variables and 2 equations, so there is (in general) one degree of freedom.(b) Differentiation gives 0 = lP dy + L′(r)dr and S ′

ydy + S ′rdr + S ′

gdg = I ′ydy + I ′

rdr . We find

dy

dg= −L′(r)S ′

g

L′(r)(S ′y − I ′

y) − lP (S ′r − I ′

r ),

dr

dg= lP S ′

g

L′(r)(S ′y − I ′

y) − lP (S ′r − I ′

r )

10. Differentiating while putting dp2 = dm = 0, gives: (i) U ′′11 dx1 + U ′′

12 dx2 = p1dλ + λdp1;(ii) U ′′

21 dx1 + U ′′22 dx2 = p2dλ; (iii) p1 dx1 + dp1x1 + p2 dx2 = 0. Solving for dx1, we obtain

∂x1

∂p1= λp2

2 + x1(p2U′′12 − p1U

′′22)

p21U

′′22 − 2p1p2U

′′12 + p2

2U′′11


2. ∂z/∂t = G′1(u, v)φ′

1(t, s) and ∂z/∂s = G′1(u, v)φ′

2(t, s) + G′2(u, v)ψ ′(s)

4. dX/dN = g(u) + g′(u)(ϕ′(N) − u

), where u = ϕ(N)/N ,

d2X/dN2 = (1/N)g′′(u)(ϕ′(N) − u

)2 + g′(u)ϕ′′(N)

6. Differentiating each side w.r.t. x while holding y constant gives 3x2 ln x + x2 = (6z2 ln z + 2z2)z′1.

When x = y = z = e, this gives z′1 = 1/2. Differentiating a second time gives 6x ln x + 5x =

(12z ln z+10z)(z′1)

2 + (6z2 ln z+2z2)z′′11. When x = y = z = e and z′

1 = 1/2, this gives z′′11 = 11/16e.

8. (a) MRS = Ryx = 2y/3x (b) MRS = Ryx = y/(x + 1) (c) MRS = Ryx = (y/x)3

10. Since y/x = (Ryx)1/3, σyx = ElRyx

(y/x) = 1/3.

12. (a) 1 (b) k (c) 0



14. Differentiate f (tx1, . . . , txn) = g(t)f (x1, . . . , xn) w.r.t. t and put t = 1, as in the proof of Euler’stheorem (Theorem 12.7.1). This yields

∑ni=1 xif

′i (x1, . . . , xn) = g′(1)f (x1, . . . , xn). Thus, by Euler’s

theorem, f must be homogeneous of degree g′(1). In fact, g(t) = tk where k = g′(1).

16. (a) Differentiating, then gathering all terms in dp and dL on the left-hand side, yields

(i) F ′(L) dp + pF ′′(L) dL = dw (ii) F(L) dp + (pF ′(L) − w) dL = L dw + dB

Since we know that pF ′(L) = w, (ii) implies that dp = (Ldw + dB)/F (L). Substituting this into (i)and solving for dL, we obtain dL = [(F (L) − LF ′(L))dw − F ′(L)dB]/pF(L)F ′′(L). It follows that

∂p

∂w= L

F(L),

∂p

∂B= 1

F(L),

∂L

∂w= F(L) − LF ′(L)

pF(L)F ′′(L),

∂L

∂B= − F ′(L)

pF(L)F ′′(L)

(b) We know that p > 0, F ′(L) > 0, and F ′′(L) < 0. Also, F(L) = (wL + B)/p > 0. Hence, it isclear that ∂p/∂w > 0, ∂p/∂B > 0, and ∂L/∂B > 0.

To find the sign of ∂L/∂w, we need the sign of F(L)−LF ′(L). From the equations in the model, weget F ′(L) = w/p and F(L) = (wL + B)/p, so F(L) − LF ′(L) = B/p > 0. Therefore ∂L/∂w < 0.


C H A P T E R 1 3 M U L T I V A R I A B L E O P T I M I Z A T I O N 53

Chapter 13 Multivariable Optimization

Chapter 13 is devoted to multivariable optimization. The first five sections treat the case of only two variables.Section 13.1 begins with a graphical explanation of the first-order conditions stated in Theorem 13.1.1.

These are first applied to a simple problem without any particular economic interpretation, and then to simpleprofit maximization in Examples 2–4. We believe the economic interpretations of conditions (∗∗) in Example3 are worth the student’s serious attention. Example 5 looks at a monopolist who sells his product in twoseparate markets. Again, the economic interpretations of the first-order conditions are important.

In identifying global maxima or minima, the theory of concave or convex functions provides sufficientconditions that are enormously important in economic analysis. Theorem 13.2.1 gives the precise result forfunctions of two variables. It is applied in Example 5 to a cost minimization problem where one of threevariables has to be eliminated.

Stationary points can be local or global maxima, local or global minima, or saddle points. Section 13.3discusses ways of distinguishing these different possibilities using second-order conditions that are formallystated in Theorem 13.3.1. (The proof is indicated at the end of the section.) Several examples and manyproblems are supplied.

Probably the most important application of multivariable calculus to economic theory is as a tool forsolving optimization problems. So interesting economic examples abound. Of these, Section 13.4 presents asmall selection, including some simple models of discriminating monopolists and monopsonists.

Section 13.5 begins with a brief discussion of some essential topological concepts in the plane—namely,interior points, open sets, boundary points, closed sets, bounded sets, and compact sets. It is pointed out thatclosed sets are typically determined by weak inequalities applied to continuous functions, and open sets bystrict inequalities applied to such functions. The extreme-value theorem claims that a continuous functiondefined on a closed and bounded set in the plane has both a maximum and minimum. This result was statedfor functions of one variable in Theorem 8.4.1. The corresponding statement for functions of two variables isgiven in Theorem 13.5.1. Frame (1) outlines a typical procedure for finding maximum and minimum values.It pays careful attention to the possibility that these may occur at boundary points, by insisting on findingmaxima and minima on the boundary as well as in the interior. Examples 1 and 2 illustrate the procedure.

Section 13.6 discusses the extension of results in earlier sections to functions of n variables. It beginswith formal definitions of maxima and minima that must be clearly understood. Topology in �n is brieflymentioned; the main difference from �2 is that the word “circle” (or “disk”) must be replaced by “ball”in any definition. Only the most essential concepts receive attention, however. Theorem 13.6.3 states theinvariance of any maximum point to (strictly) increasing transformations of the maximand. This result oftengets overlooked despite its enormous importance and usefulness in economics. Indeed, the different roleof strictly increasing and merely increasing transformations, illustrated in Problem 2, should also be notedcarefully. Finally, Section 13.7 discusses the envelope theorem which appears so often in economics. Problem4 is typical of the way in which the envelope theorem can give interesting economic conclusions which arefar from evident to the uninitiated, and impossible even to understand without some facility in multivariablecalculus.


54 C H A P T E R 1 3 M U L T I V A R I A B L E O P T I M I Z A T I O N


13.12. (a) (x, y) = (3, −4). (f ′

1(x, y) = 2x − 6 and f ′2(x, y) = 2y + 8, so the only stationary point is

(x, y) = (3, −4).) (b)f (x, y) = x2−6x+32+y2+8y+42+35−32−42 = (x−3)2+(y+4)2+10 ≥ 10for all (x, y), whereas f (3, −4) = 10, so (3, −4) minimizes f .

4. (a) P(10, 8) = P(12, 10) = 98 (b) (x, y) = (11, 9) maximizes P , and P(11, 9) = 100.

13.22. (a) Profit: π(x, y) = 24x + 12y − C(x, y) = −2x2 − 4y2 + 4xy + 64x + 32y − 514. Maximum at

x = 40, y = 24. (Then π(40, 24) = 1150.) Since π ′′11 = −4 ≤ 0, π ′′

22 = −8 ≤ 0, and π ′′11π

′′22 −

(π ′′12)

2 = 16 ≥ 0, we have found the maximum. (b) x = 34, y = 20. (With y = 54 − x, profits areπ = −2x2 − 4(54 − x)2 + 4x(54 − x) + 64x + 32(54 − x) − 514 = −10x2 + 680x − 10450, whichhas a maximum at x = 34. Then y = 54 − 34 = 20. The maximum value is 1110.)

4. (a) π(x, y) = px + qy − C(x, y) = (25 − x)x + (24 − 2y)y − (3x2 + 3xy + y2) = −4x2 − 3xy −3y2 + 25x + 24y. (b) π ′

1 = −8x − 3y + 25 = 0 and π ′2 = −3x − 6y + 24 = 0 when (x, y) = (2, 3).

Moreover, then π ′′11 = −8 ≤ 0, π ′′

22 = −6 ≤ 0, and π ′′11π

′′22 − (π ′′

12)2 = (−8)(−6) − (−3)2 = 39 ≥ 0.

So (x, y) = (2, 3) maximizes profits.

6. (a) The first-order conditions are: π ′1 = p−2αx = 0, π2 = q−2βy = 0. Thus the only stationary point is

x∗ = p/2α, y∗ = q/2β. Moreover, π ′′11 = −2α ≤ 0, π ′′

22 = −2β ≤ 0, and π ′′11π

′′22 − (π ′′

12)2 = 4αβ ≥ 0,

so (x∗, y∗) = (p/2α, q/2β) maximizes profits.(b) π∗(p, q) = px∗ + qy∗ − α(x∗)2 − β(y∗)2 = p2/4α + q2/2β. Hence ∂π∗(p, q)/∂p = p/2α = x∗:Increasing the price p by one unit increases the optimal profit by x∗, the optimal production level of thefirst good. ∂π∗(p, q)/∂q = y∗ has a similar interpretation.

8. The second-order partial derivatives are f ′′11 = a(a − 1)Axa−2yb, f ′′

12 = f ′′21 = abAxa−1yb−1, and

f ′′22 = b(b−1)Axayb−2. Thus, f ′′

11f′′22−(f ′′

12)2 = abA2x2a−2y2b−2

[1−(a+b)

]. Suppose that a+b ≤ 1.

Then a ≤ 1 and b ≤ 1 as well. If x > 0 and y > 0, then f ′′11 ≤ 0 and f ′′

22 ≤ 0, and f ′′11f

′′22 − (f ′′

12)2 ≥ 0.

We conclude from Note 13.2.2 that f is concave for x > 0, y > 0 when a + b ≤ 1.

13.32. (a) f ′

1 = 2x + 2y2, f ′2 = 4xy + 4y, f ′′

11 = 2, f ′′12 = 4y, f ′′

22 = 4x + 4(b) f ′

2 = 0 ⇐⇒ 4y(x + 1) = 0 ⇐⇒ x = −1 or y = 0. If x = −1, then f ′1 = 0 for y = ±1. If

y = 0, then f ′1 = 0 for x = 0. Thus we get the three stationary points classified in the table:

(x, y) A B C AC − B2 Type of stationary point:

(0, 0) 2 0 4 8 Local minimum point

(−1, 1) 2 4 0 −16 Saddle point

(−1, −1) 2 −4 0 −16 Saddle point



4. (a) f ′1(x, y) = 24x2 − 12y = 12(2x2 − y), f ′

2(x, y) = −12x + 3y2 = 3(y2 − 4x). The stationarypoints are thus the solutions of the system (i) 2x2 − y = 0, (ii) −4x + y2 = 0. From (i) we get y = 2x2

which inserted into (ii) yields 4x(x3 − 1) = 0. The only solutions of this equation are x = 0 and x = 1.The stationary points are thus (0, 0) and (1, 2). Moreover, f ′′

11(x, y) = 48x, f ′′12(x, y) = −12, and

f ′′22(x, y) = 6y. The stationary points are classified in the table:


(0, 0) 0 −12 0 −144 Saddle point

(1, 2) 48 −12 12 432 Local minimum point

(b) Since g(x, y) is a square, g(x, y) ≥ 0 for all x and y. On the other hand, g(0, 0) = 0, so (0, 0) isa global minimum point for g. (There are many global minimum points for g, namely all points (x, y)

satisfying 8x3 − 12xy + y3 = 0.)

6. In all three cases we easily find that (0, 0) is a stationary point where z = 0 and A = B = C = 0, soAC −B2 = 0. In case (a), z ≤ 0 for all (x, y), so the origin is a maximum point. In case (b), z ≥ 0 for all(x, y), so the origin is a minimum point. In (c), z takes positive and negative values at points arbitrarilyclose to the origin, so it is a saddle point.

8. (a) f is defined for x = 0 and for y > −1/x2. (b) f ′1(x, y) = 2xy

1 + x2yand f ′

2(x, y) = x2

1 + x2y. Here

f ′1 = f ′

2 = 0 at all points (0, b) with b ∈ �. (c) Because AC − B2 = 0 when (x, y) = (0, b), thesecond-derivative test fails. But by studying the function directly, we see that (0, b) is a local maximumpoint if b < 0; a saddle point if b = 0; and a local minimum point if b > 0. See Fig. M13.3.8.

x

y

z

x

y

z

z = ln(1 + x2y)

Figure M13.3.8

13.4

2. (a) π = −bp2 − dq2 + (a + βb)p + (c + βd)q − α − β(a + c), p∗ = a + βb

2b, q∗ = c + βd

2d.

The second-order conditions are obviously satisfied because π ′′11 = −2b, π ′′

12 = 0, and π ′′22 = −2d.

(b) The new profit function is π = −bp2 − dp2 + (a + βb)p + (c + βd)p − α − β(a + c) and the price

which maximizes profits is p = a + c + β(b + d)

2(b + d).



(c) If β = 0, then p∗ = a

2b, q∗ = c

2d, and π(p∗, q∗) = a2

4b+ c2

4d− α. Moreover, p = a + c

2(b + d)with

π(p) = (a + c)2

4(b + d)− α, and π(p∗, q∗) − π(p) = (ad − bc)2

4bd(b + d)≥ 0. Note that the difference is 0 when

ad = bc, in which case p∗ = q∗, so the firm wants to charge the same price in each market anyway.

4. (a) Let (x0, y0) = (0, 11.29), (x1, y1) = (1, 11.40), (x2, y2) = (2, 11.49), and (x3, y3) = (3, 11.61),so that x0 corresponds to 1970, etc. (The numbers yt are approximate, as are most subsequent results.)We find that µx = 1

4 (0 + 1 + 2 + 3) = 1.5, µy = 14 (11.29 + 11.40 + 11.49 + 11.61) = 11.45,

and σxx = 14 [(0 − 1.5)2 + (1 − 1.5)2 + (2 − 1.5)2 + (3 − 1.5)2] = 1.25. Moreover, we find σxy =

14 [(−1.5)(11.29−11.45)+(−0.5)(11.40−11.45)+(0.5)(11.49−11.45)+(1.5)(11.61−11.45)], which

is equal to 0.13125, and so a = σxy/σxx = 0.105 and b = µy − aµx ≈ 11.45 − 0.105 · 1.5 = 11.29.

(b) With z0 = ln 274, z1 = ln 307, z2 = ln 436, and z3 = ln 524, we have (x0, z0) = (0, 5.61),(x1, z1) = (1, 5.73), (x2, z2) = (2, 6.08), and (x3, z3) = (3, 6.26). As before, µx = 1.5 and σxx = 1.25.Moreover, µz = 1

4 (5.61 + 5.73 + 6.08 + 6.26) = 5.92 and

σxz ≈ 14 [(−1.5)(5.61−5.92)+(−0.5)(5.73−5.92)+(0.5)(6.08−5.92)+(1.5)(6.26−5.92)] = 0.2875

Hence c = σxz/σxx = 0.23, d = µz − cµx = 5.92 − 0.23 · 1.5 = 5.575.

(c) With ln (GNP) = 0.105x + 11.25, GNP = e11.25e0.105x = 80017e0.105x . Likewise, FA = 256e0.23x .The requirement that FA = 0.01 GNP implies that e0.23x−0.105x = 80017/25600, and so 0.125x =ln(80017/25600). Thus x = ln(80017/25600)/0.125 = 9.12. Since x = 0 corresponds to 1970, thegoal would have been reached in 1979.

13.5

2. (a) Stationary points occur where both (i) f ′1(x, y) = 3x2 − 9y = 0 and (ii) f ′

2(x, y) = 3y2 − 9x = 0.From (ii) we get x = 1

3y2, which inserted into (i) yields y4 = 27y, whose only solutions are y = 0 andy = 3. So the only interior stationary point is (3, 3). Along the edges x = 0 and y = 0 of the boundary,the function has no extreme point except at a corner. Along the edges x = 4 and y = 4, we find twocandidates for extreme points at (4, 2

√3) and (2

√3, 4). There are also the four corners (0, 0), (4, 0),

(0, 4), and (4, 4). Comparing the values of f at all these points, we find that f has maximum 91 at (0, 4)

and at (4, 0), and minimum 0 at (3, 3). (Extreme points exist by the extreme-value theorem.)

(b) At any stationary point, f ′1 = 2x − 1 = 0 and f ′

2 = 4y = 0, so f is stationary at the interior point( 1

2 , 0) of the domain. Along the boundary x2 + y2 = 1, the behaviour of f is described by the functiong(x) = x2 + 2(1 − x2) − x = 2 − x − x2, with x in [−1, 1]. We see that g′(x) = −1 − 2x = 0 atx = − 1

2 , so the points (− 12 , ± 1

2

√3) are optimality candidates. Moreover, g(−1) = 2 and g(1) = 0,

so (−1, 0) and (1, 0) are also optimality candidates. Comparing the values of f at all these points, weconclude that f has maximum 9/4 at (−1/2,

√3/2) and at (−1/2, −√

3/2). The minimum is −1/4 at(1/2, 0).

(c) At any stationary point, f ′1 = 3x2 − 2x = 0 and f ′

2 = −2y = 0, so f is stationary at ( 23 , 0) (and at

the boundary point (0, 0)). Along the edge x = 0, y ∈ [−1, 1], f (0, y) = 3 − y2, which is largest aty = 0 and smallest at y = ±1. Along the edge x2 +y2 = 1, x ∈ [0, 1], one has f (x, y) = 2 +x3, whichis strictly increasing, smallest at x = 0, largest at x = 1. Comparing the values of f at all these points,we conclude that the maximum is 3 at (0, 0) and at (1, 0). The minimum is 2 at (0, −1) and at (0, 1).



(d) f ′1(x, y) = (2x2 − 5x + 3)ex2−x(2y − 1)e(y−2)2

and f ′2(x, y) = 2(2y2 − 5y + 3)(x − 2)ex2−xe(y−2)2

,so the function f has five stationary points, with (x, y) = (1, 1), ( 3

2 , 1), (1, 32 ), ( 3

2 , 32 ), or (2, 1

2 ). Noneof these are in the interior of the domain, however. So any maximum or minimum must be on one of thefour edges, or at the four corners. Obviously f = 0 along the edges x = 2 and y = 1

2 , as well as at thethree corners which make up the ends of these edges. At the other corner, f (0, 0) = 2e4. Along the edgex = 0 one has f ′

2 < 0 for all y in [0, 12 ], so only the corners are relevant. It remains to consider the edge

y = 0, along which f ′1 < 0 for 0 ≤ x < 1 and for 3

2 < x ≤ 2, but f ′1 > 0 for 1 < x < 3

2 . Note thatf (1, 0) = e4 > 0 and f ( 3

2 , 0) = 12e19/4 < 2e4. We conclude that the maximum is 2e4 at (0, 0) and the

minimum is 0 at (x, 1/2) for all x ∈ [0, 2], and at (2, y) for all y ∈ [0, 1/2].

4. (a) The first-order conditions 2axy + by + 2y2 = 0 and ax2 + bx + 4xy = 0 must have (x, y) =(2/3, 1/3) as a solution. So a = 1 and b = −2. Also c = 1/27, so that f (2/3, 1/3) = −1/9. BecauseA = f ′′

11(2/3, 1/3) = 2/3, B = f ′′12(2/3, 1/3) = 2/3, and C = f ′′

22(2/3, 1/3) = 8/3, Theorem 13.3.1shows that this is a local minimum.

(b) Maximum 193/27 at (2/3, 8/3). Minimum −1/9 at (2/3, 1/3). (The extreme points must be stationarypoints in the interior of the admissible set or else points on the boundary. Note that we must check ifthere are stationary points in the interior other than (2/3, 1/3). The first-order conditions reduce to2y(x − 1 + y) = 0 and x(x − 2 + 4y) = 0. The stationary points satisfying these equations are (0, 0),(0, 1), (2, 0), and (2/3, 1/3). Three of these four are on the boundary. The maximum is a point on theboundary line 2x + y = 4.)

6. (a) Closed and bounded, so compact. (b) Open and unbounded. (c) Closed and bounded, so compact.(d) Closed and unbounded. (e) Closed and unbounded. (f) Open and unbounded.

8. (a) The domain D is shown in Fig. M13.5.8. The first order partials aref ′

1(x, y) = e− 14 (x+y2)

(1 − 1

4 (x + y)), f ′

2(x, y) = e− 14 (x+y2)

(1 − 1

2y(x + y)).

(b) The unique stationary point (7/2, 1/2) gives the maximum value 4e−15/16.

y

x

y = 1 − x

D

Figure M13.5.8

13.62. (a) f (x) = e−x2

and g(x) = F(f (x)) = ln(e−x2) = −x2 both have a unique maximum at x = 0.

(b) Only x = 0 maximizes f (x). But g(x) = 5 is maximized at every point x because it is a constant.

4. f ′x = −6x2 + 30x − 36, f ′

y = 2 − ey2, f ′

z = −3 + ez2. The 8 stationary points are (x, y, z) =

(3, ±√ln 2, ±√

ln 3 ), and (x, y, z) = (2, ±√ln 2, ±√

ln 3 ), where all combinations of signs are allowed.

13.72. (a) K∗ = 8

27p3r−3, L∗ = 14p2w−2, T ∗ = 1

3√

3p3/2q−3/2

(b) Q∗ = 49p2r−2 + 1

2pw−1 + 1√3p1/2q−1/2, so ∂Q∗/∂r = − 8

9p2r−3 = −∂K∗/∂p



4. (a) First-order conditions: (i) R′1 − C ′

1 + s = 0, (ii) R′2 − C ′

2 − t = 0.(b) π ′′

11 = R′′11 − C ′′

11 < 0 and D = π ′′11π

′′22 − (π ′′

12)2 = (R′′

11 − C ′′11)(R

′′22 − C ′′

22) − (R′′12 − C ′′

12)2 > 0.

(c) Taking the total differentials of (i) and (ii) yields

(R′′11 − C ′′

11)dx∗1 + (R′′

12 − C ′′12)dx∗

2 = −ds, (R′′21 − C ′′

21)dx∗1 + (R′′

22 − C ′′22)dx∗

2 = dt

Solving for dx∗1 and dx∗

2 yields, after rearranging,

dx∗1 = −(R′′

22 − C ′′22)ds − (R′′

12 − C ′′12)dt

D, dx∗

2 = (R′′21 − C ′′

21)ds + (R′′11 − C ′′

11)dt

D

From this we find that the partial derivatives are

∂x∗1

∂s= −R′′

22 + C ′′22

D> 0,

∂x∗1

∂t= −R′′

12 + C ′′12

D> 0,

∂x∗2

∂s= R′′

21 − C ′′21

D< 0,

∂x∗2

∂t= R′′

11 − C ′′11

D< 0

where the signs follow from the assumptions in the problem and the fact that D > 0 from (b). Notethat these signs accord with economic intuition. For example, if the tax on good 2 increases, then theproduction of good 1 increases, while the production of good 2 decreases.(d) Follows from the expressions in (c) because R′′

12 = R′′21 and C ′′

12 = C ′′21.


2. (a) The profit is π(Q1, Q2) = 120Q1 + 90Q2 − 0.1Q21 − 0.1Q1Q2 − 0.1Q2

2. The first-order conditionsgive Q1 = 500, Q2 = 200. We easily see that the second-order conditions are satisfied.(b) The profit is now π(Q1, Q2) = P1Q1 + 90Q2 − 0.1Q2

1 − 0.1Q1Q2 − 0.1Q22. The first-order

conditions give π ′1 = P1 − 0.2Q1 − 0.1Q2 = 0 and π ′

2 = 90 − 0.1Q1 − 0.2Q2 = 0. If we requireQ1 = 400, then P1 − 80 − 0.1Q2 = 0 and 90 − 40 − 0.2Q2 = 0. It follows that Q2 = 250 and soP1 = 105. Second-order conditions are easily checked.

4. (a) f ′1 = 2(x + y − 2) + 4x(x2 + y − 2), f ′

2 = 2(x + y − 2) + 2(x2 + y − 2), f ′′11 = 12x2 + 4y − 6,

f ′′12 = f ′′

21 = 4x+2, f ′′22 = 4. (b) f ′

1 = f ′2 = 0 implies that x+y−2 = −2x(x2+y−2) = −(x2+y−2).

The last equality gives x = 12 or y = 2 − x2. Inserting each of these into f ′

2 = 0 yields three stationarypoints, which are classified in the following table.




(1/2, 13/8) 7/2 4 4 −2 Saddle point

(c) f (0, 2) = f (1, 1) = 0, and f (x, y) ≥ 0 for all (x, y), so these are (global) minimum points.



6. (a) f is stationary when 3x2 − 2xy = −x2 + 2y = 0. Substituting 2y = x2 in the first expression yields3x2 = x3, implying that x = 0 or x = 3. There are two stationary points at (0, 0 and (3, 9/2).(b) (0, 0), ( 1

2

√2,

√2), (− 1

2

√2, −√

2). (The stationary points must satisfy y(8x2 − 5xy + 1) = 0 andx(2y2 − 5xy + 1) = 0. Notice that at any stationary point, x = 0 implies y = 0, and conversely. Soapart from (0, 0), any other stationary point must satisfy 5xy − 1 = 8x2 = 2y2. Since no solution ofthese last two equations can satisfy y = −2x, the only possibility is that y = 2x and so 2x2 = 1.)

8.(

f ′′11 f ′′

12

f ′′21 f ′′

22

)=( 1

2e−x−y − e−x 12e−x−y

12e−x−y 1

2e−x−y − e−y

).

It follows that f ′′11 = e−x( 1

2e−y − 1) < 0 because e−y < 1 < 2. In the same way, f ′′22 < 0. Moreover,

f ′′11f

′′22 − (f ′′

12)2 = 1

2e−x−y(2 − e−x − e−y) > 0, so f is concave.

10. (a) f ′1 = f ′

2 = 0 implies that 2x − y − 3x2 = −2y − x = 0. So y = − 12x and 5

2x − 3x2 = 0, implyingthat x = 0 or 5

6 . There are two stationary points, which are classified in the following table.


(0, 0) 2 −1 −2 −5 Saddle point

(5/6, −5/12) −3 −1 −2 5 Local maximum point

(b) f ′′11 = 2 − 6x ≤ 0 ⇐⇒ x ≥ 1/3, while f ′′

22 = −2 ≤ 0, and f ′′11f

′′22 − (f ′′

12)2 = 12x − 5 ≥ 0 ⇐⇒

x ≥ 5/12. We conclude that f is concave if and only if x ≥ 5/12. The largest value of f in S is 125/432,obtained at (5/6, −5/12).


60 C H A P T E R 1 4 C O N S T R A I N E D O P T I M I Z A T I O N

Chapter 14 Constrained Optimization

Constrained optimization is the central tool of mathematical economics. Economics is about making the bestuse of scarce resources, and scarcity is naturally represented by constraints. The usual technique for solvingconstrained optimization problems is the Lagrange multiplier method. Most of this chapter is concerned withequality constraints; the last two sections offer a brief introduction to inequality constraints.

For the case of two variables, the Lagrange multiplier method is the subject of Section 14.1. The methodis set out as a procedure in a large frame that should be memorized by the students. Partly because it easilyleads to errors when there are inequality constraints, we prefer not to set to zero the partial derivative ofthe Lagrangian w.r.t. the multiplier (as is done in many other textbooks).2 Example 2 is important becauseit explains what needs to be done when the equation system has multiple solutions. The consumer demandproblem with a Cobb–Douglas utility function is considered in Example 3, and the appropriate demandfunctions are derived.

Section 14.2 presents the standard economic interpretation of Lagrange multipliers as shadow prices.For the case when the (optimal) value function is differentiable, we present an easy proof that its derivativeis equal to the Lagrange multiplier. Example 1 illustrates its use in economics. The section ends with a richselection of problems.

Section 14.3 is devoted to explaining first with a geometric argument why the Lagrangian method works—when it does. Then an analytic argument for the first-order conditions is given, and a precise result is formulatedin Theorem 14.3.1. Example 1 considers a utility maximization problem in which a mechanical applicationof the Lagrangian method fails to give the right answer. Problem 1 shows that the solution to a constrainedmaximization problem need not maximize the Lagrangian, even though (under usual conditions) it must makethe Lagrangian stationary. This is another point that is rather commonly misunderstood. Let the teacher bewareof misleading attempts to motivate the Lagrangian technique! Fortunately, the error found in the quotation inProblem 2 is far less common.

In Section 14.4 we turn from necessary to sufficient conditions for optimality. Of these, the simplest tostate is that a global maximum (or minimum) of the Lagrangian which satisfies the constraint is optimal. Thiscould have been stated as a theorem in its own right. In practice, however, there are difficulties in verifying thata global maximum (or minimum) has been found, except in the important case when the Lagrangian happensto be concave (or convex). This is the case mentioned in Theorem 14.4.1. (The appropriate generalization toquasi-concave functions is not, in our view, an “essential” topic for an undergraduate course.)

The concept of a local constrained extremum is introduced. Theorem 14.4.2 then gives sufficient condi-tions for the Lagrangian method to find a local constrained extremum. In smaller print the sufficient conditionsin Theorem 14.4.2 are expressed in terms of the sign of the bordered Hessian determinant.

Sections 14.1–14.4 consider only functions of two variables. The extension to n variables in Section14.5 is mostly routine, though actually solving problems with many variables is often very complicated.Example 3 discusses how to derive individual demand functions in a general n good setting. Then the caseof m equality constraints is considered. The Lagrangian needs as many Lagrange multipliers as there areconstraints, of course. This gives the right number of equations when the first-order conditions are combinedwith the constraints themselves. In Example 5 we study a welfare maximization problem in a simple exchangeeconomy. Problem 3 is an interesting economic example, involving the optimal choice of leisure as well as

2 The equality constraint is an obvious necessary condition for optimality. There is no reason to differentiateanything to demonstrate this.


C H A P T E R 1 4 C O N S T R A I N E D O P T I M I Z A T I O N 61

consumption of two goods. It also shows the possibility of a corner solution with zero labour supply in caseunearned income is sufficiently high.

Section 14.6 deals with sensitivity analysis. First, the standard interpretation of Lagrange multipliersas shadow prices is mentioned. With m constraints, the value function depends on m variables. Where it isdifferentiable, the m Lagrange multipliers are partial derivatives of the value function. Example 1 verifies theproperties of Lagrange multipliers numerically for a case with two constraints.

Adjusting the right-hand side of an equality constraint is just one of many perturbations to which aconstrained optimization problem can be subjected. Other perturbations are considered in this section; ineconomics they feature constantly in comparative static propositions, of course. For such perturbations, frame(6) states one version of the “envelope theorem”, which often arises in economic analysis. Examples 2, 3, and4 involve interesting economic applications.

The final Sections 14.7 and 14.8 give a brief introduction to nonlinear programming—i.e., optimizationproblems where the contraints are expressed as inequalities, rather than equalities. For the case of two variablesand one constraint, a frame sets out a recipe for solving the problem based on the Lagrange multiplier method.Part of the recipe involves a complementary slackness condition. Theorem 14.7.1 emphasizes how this methodleads to a sufficient condition when the Lagrangian is concave. Section 14.7 concludes with a graphicalexplanation of the complementary slackness condition.

In Section 14.8, the recipe of Section 14.7 is extended to general problems with n variables and m

constraints. Particular attention is paid to nonnegativity restrictions on the variables. The Lagrange multipliersare related to partial derivatives of the (optimal) value function.


14.12. (a) (x, y) = (4/5, 8/5) with λ = 8/5. (With L(x, y) = x2+y2−λ(x+2y−4), the first-order conditions

are L′1 = 2x − λ = 0 and L′

2 = 2y − 2λ = 0. From these equations we get 2x = y, which inserted intothe constraint gives x + 4x = 4. So x = 4/5 and y = 2x = 8/5, with λ = 2x = 8/5.)(b) The same method as in (a) gives 2x − λ = 0 and 4y − λ = 0, so x = 2y. From the constraint we getx = 8 and y = 4, with λ = 16.

4. (a) With L = x2 + y2 − 2x + 1 − λ(x2 + 4y2 − 16), the first-order conditions are 2x − 2 − 2λx = 0 and2y−8λy = 0. One possibility is λ = 1/4, in which case x = 4/3 and so y = ±4

√2/3. Alternatively, y =

0, in which case x = ±4. So there are four solution candidates: (i) (x, y, λ) = (4, 0, 3/4), (ii) (x, y, λ) =(−4, 0, −5/4), (iii) (x, y, λ) = (4/3, 4

√2/3, 1/4), and (iv) (x, y, λ) = (4/3, −4

√2/3, 1/4). Of these,

the second is the maximum point (while (iii) and (iv) are the minimum points).(b) The first-order conditions imply that 2x + 3y = λ = 3x + 2y. So (x, y) = (50, 50) with λ = 250.

6. (a) With L(x, y) = 100 − e−x − e−y − λ(px + qy − m), L′x = L′

y = 0 when e−x = λp and e−y = λq.Hence, x = − ln(λp) = − ln λ − ln p, y = − ln λ − ln q. Inserting these expressions for x and y

into the constraint, then solving for ln λ, yields ln λ = −(m + p ln p + q ln q)/(p + q). Thereforex(p, q, m) = [m + q ln(q/p)]/(p + q), y(p, q, m) = [m + p ln(p/q)]/(p + q).(b) x(tp, tq, tm) = [tm + tq ln(tq/tp)]/(tp + tq) = x(p, q, m), so x is homogeneous of degree 0. Inthe same way we see that y(p, q, m) is homogeneous of degree 0.

14.22. (a) L(x, y) = x +y −λ(x2 +y −1). The equations L′

1 = 1−2λx = 0, L′2 = 1−λ = 0, and x2 +y = 1



have the solution x = 1/2, y = 3/4, and λ = 1. (b) The solution is illustrated in Fig. M14.2.2. Theminimization problem has no solution because f (x, 1 − x2) = x + 1 − x2 → −∞ as x → ∞.(c) The new solution is x = 0.5 and y = 0.85. The change in the value function is f ∗(1.1) − f ∗(1) =(0.5 + 0.85) − (0.5 + 0.75) = 0.1. Because λ = 1, one has λ · dc = 1 · 0.1 = 0.1. So, in this case, (3)is satisfied with equality. (This is because of the special form of the functions f and g.)

y

x−1 1

y = 1 − x2( 1

2 , 34 )

x + y = 12

x + y = 54

y

−6−5

−4−3−2−1

12

3456

x−3−2−1 1 2 3 4

(x, y)

y2 = x(x + 1)2

d


4. (a) x∗ = 4, y∗ = 24, λ = 1/4. (b) y = 97/4, x = 4. �U = 105/4 − 104/4 = 1/4, the value of theLagrange multiplier from (a). (There is exact equality here because U is linear in one of the variables.)(c) x∗ = q2/4p, y∗ = m/q − q/4p. (Note that y∗ > 0 iff m > q2/4p.)

14.32. The problem with systems of three equations and two unknowns is not that they are merely difficult to

solve but that they are usually inconsistent—i.e., it is impossible to solve them. The equations f ′x(x, y) =

f ′y(x, y) = 0 are NOT valid at the optimal point.

4. The minimum is 1 at (x, y) = (−1, 0). Actually, this problem is quite tricky. The Lagrangian can bewritten in the form L = (x + 2)2 + (1 − λ)y2 + λx(x + 1)2. The only stationary point which satisfiesthe constraint is (0, 0), with λ = −4, and with f (0, 0) = 4. (In fact, L′

2 = 0 only if λ = 1 or y = 0. Forλ = 1, L′

1 = 3(x + 1)2 + 2 > 0 for all x. For y = 0, the constraint gives x = 0 or x = −1. But x = −1gives L′

1 = 2, so x = 0 is necessary for a stationary point.) Yet at (−1, 0) both g′1(−1, 0) and g′

2(−1, 0)

are 0, and the Lagrange multiplier method fails. The given problem is to minimize (the square of) thedistance from (−2, 0) to a point on the graph of g(x, y) = 0. But the graph consists of the isolated point(−1, 0) and a smooth curve, as illustrated in Fig. M14.3.4.

14.42. With L(x, y) = x2 + y2 − λ(x + 2y − a) the first-order conditions are 2x − λ = 0 = 2y − 2λ.

So (x, y) = (a/5, 2a/5) with λ = 2a/5. Applying Theorem 14.4.2 with f (x, y) = x2 + y2 andg(x, y) = x + 2y, we find that D(x, y, λ) = 10, so the only stationary point is a local minimum. (Byreducing the problem to a one-variable problem, it is easy to see that it is actually a global minimum.)Geometrically, the problem is to find that point on the straight line x +2y = a that is closest to the origin.



14.5

2. Here L = x + 4y + 3z − λ(x2 + 2y2 + 13z2 − b). So necessary conditions are:

(i) L′1 = 1 − 2λx = 0; (ii) L′

2 = 4 − 4λy = 0; (iii) L′3 = 3 − 2

3λz = 0. It follows that λ �= 0,and so x = 1/2λ, y = 1/λ, z = 9/2λ. Inserting these values into the constraint yields λ2 = 9/b, soλ = ±3/

√b. The value of the objective function is x + 4y + 3z = 18/λ, so λ = −3/

√b determines the

minimum point. This is (x, y, z) = (a, 2a, 9a), where a = −√b/6.

4. The constraints reduce to h + 2k + � = 0 and 2h − k − 3� = 0, so k = −h and � = h. But thenf (x, y, z) = x2 + y2 + z2 = 200 + 3h2 ≥ 200 for all h, so f is maximized for h = 0. Then k = � = 0also, and we conclude that (x, y, z) = (10, 10, 0) solves the minimization problem.

6. (a) ci1 = α1r

i/p1, ci2 = α2r

i/p2. (Write the utility function in the equivalent form Ui(ci1, c

i2) =

ln[(ci

1)α1(ci

2)α2]. Using Theorem 13.6.3, the result follows immediately from (∗∗) in Example 14.1.3,

recalling that α1 + α2 = 1.) (b) In Example 5, λ1 = α1/e1, λ2 = α2/e2. So p1 = λ1 and p2 = λ2

implies ci1 = α1r

i/p1 = rie1 = βie1 provided that ri = βi , which also gives ci2 = βie2. Because

p1e1 + p2e2 = λ1e1 + λ2e2 = α1 + α2 = 1, it follows that ri = βi(p1e1 + p2e2) for i = 1, 2.

8. (a) With a Cobb–Douglas utility function, U ′k(x) = αkU(x)/xk , so from (6) (with j = 1), we have

pk/p1 = U ′k(x)/U ′

1(x) = αkx1/α1xk . Thus pkxk = (ak/a1)p1x1. Inserted into the budget constraint, wehave p1x1 + (a2/a1)p1x1 + · · · + (an/a1)p1x1 = m, which implies that p1x1 = a1m/(a1 + · · · + an).Similarly, pkxk = akm/(a1 + · · · + an) for k = 1, . . . , n.

(b) From (6) (with j = 1), we get xa−1k /xa−1

1 = pk/p1 and so xk/x1 = (pk/p1)−1/(1−a) or pkxk/p1x1 =

(pk/p1)1−1/(1−a) = (pk/p1)

−a/(1−a). The budget constraint gives

p1x1[1 + (p2/p1)

−a/(1−a) + · · · + (pn/p1)−a/(1−a)

] = m. So p1x1 = mp−a/(1−a)1

/ n∑i=1

p−a/(1−a)

i .

Arguing similarly for each k, one has xk = mp−1/(1−a)

k

/ n∑i=1

p−a/(1−a)

i for k = 1, . . . , n.

10. Differentiating the constraint w.r.t. x1 (keeping x2 fixed) yields g′1 + g′

3(∂x3/∂x1) = 0, which implies(i) ∂x3/∂x1 = −g′

1/g′3. Similarly, (ii) ∂x3/∂x2 = −g′

2/g′3. The first-order conditions for the maximiza-

tion of z = f (x1, x2, x3), where x3 is a function of (x1, x2), are (iii) ∂z/∂x1 = f ′1 +f ′

3(∂x3/∂x1) = 0 and(iv) ∂z/∂x2 = f ′

2 + f ′3(∂x3/∂x2) = 0. Substitute from (i) and (ii) into (iii) and (iv), letting λ = f ′

3/g′3.

This yields the equations f ′1 − λg′

1 = 0 and f ′2 − λg′

2 = 0. Also, by definition of λ, f ′3 − λg′

3 = 0. Theseare the conditions (3) for n = 3.

14.6

2. (a) x = aM/α, y = bM/β, z = cM/γ , λ = 1/(2M), where M = √L/√

a2/α + b2/β + c2/γ . (Thefirst-order conditions give x = a/2λα, y = b/2λβ, z = c/2λγ . Substituting in the constraint and solvingfor λ gives the solution.) (b) We find that M = √

L/5, and the given values of x, y, and z follow. ForL = 100 one has M = 2 and λ = 1/4. The increase in the maximal value as L increases from 100 to101, is approximately λ · 1 = 0.25. The actual increase is 5(

√101 − √

100 ) ≈ 0.249378.

4. K∗ = 21/3r−1/3w1/3Q4/3, L∗ = 2−2/3r2/3w−2/3Q4/3, C∗ = 3 · 2−2/3r2/3w1/3Q4/3,λ = 24/3r2/3w1/3Q1/3. The equalities (∗) are easily verified.



6. (a) Given L = ex + y + z − λ(x + y + z − 1) − µ(x2 + y2 + z2 − 1), the first-order conditions are(i) ex − λ − 2µx = 0; (ii) 1 − λ − 2µy = 0; (iii) 1 − λ − 2µz = 0. From (ii) and (iii) it follows thatµ(y − z) = 0. If µ = 0, then λ = 1, x = 0, and y + z = 1, y2 + z2 = 1. The stationary points occurwhere (x, y, z) = (0, 1, 0) or (0, 0, 1), in which case ex + y + z = 2. If µ �= 0, then y = z = 1

2 (1 − x),implying that x2 + 1

2 (1 − x)2 = 1, so x = 1 or − 13 . The maximum is at (1, 0, 0) where ex + y + z = e.

Another stationary point is at (− 13 , 2

3 , 23 ) where ex + y + z = e−1/3 + 4

3 < 73 < e. The relevant Lagrange

multipliers are λ = 1, µ = 12 (e − 1). (b) �f ∗ ≈ λ · (0.02) + µ · (−0.02) = 0.01(3 − e) ≈ 0.0028.

14.72. (a) The Lagrangian is L(x, y) = x2 + 2y2 − x − λ(x2 + y2 − 1). Conditions (2)–(4) take the form:

2x − 1 − 2λx = 0; 4y − 2λy = 0; λ ≥ 0 (λ = 0 if x2 + y2 < 1). (b) Candidates: (1/2, 0) withλ = 0; (1, 0) with λ = 1/2; (−1, 0) with λ = 3/2; and (−1/2, ±√

3/2) with λ = 2. Maximum 9/4 at(−1/2,

√3/2) and at (−1/2, −√

3/2). The extreme-value theorem confirms that this is the solution.

4. (a) f ′1(x, y) ≥ 0 (f ′

1(x, y) = 0 if x < b) and f ′2(x, y) = 0 (b) f ′

1(x, y) = 0 and f ′2(x, y) ≤ 0

(f ′2(x, y) = 0 if y > a)

14.82. (a) The admissible set is the shaded region in Fig. M14.8.2.

(b) With the constraints g1(x, y) = −x − y ≤ −4, g2(x, y) = −x ≤ 1, g3(x, y) = −y ≤ −1, theLagrangian is L = x + y − ex − ex+y − λ1(−x − y + 4) − λ2(−x − 1) − λ3(−y + 1). The first-orderconditions are that there exist nonnegative numbers λ1, λ2, and λ3 such that:(i) L′

x = 1 − ex − ex+y + λ1 + λ2 = 0; (ii) L′y = 1 − ex+y + λ1 + λ3 = 0; (iii) λ1(−x − y + 4) = 0;

(iv) λ2(−x − 1) = 0; (v) λ3(−y + 1) = 0. (We formulate the complementary slackness conditions asin (14.7.5).) From (ii), ex+y = 1 + λ1 + λ3. Inserting this into (i) yields λ2 = ex + λ3 ≥ ex > 0.

Because λ2 > 0, (iv) implies that x = −1. So any solution must lie on the line (II) in the figure, whichshows that the third constraint must be slack. (Algebraically, because x + y ≥ 4 and x = −1, we havey ≥ 4 − x = 5 > 1.) So from (v) we get λ3 = 0, and then (ii) gives λ1 = ex+y − 1 ≥ e4 − 1 > 0.

Thus from (iii), the first constraint is active, so y = 4 − x = 5. Hence the only possible solution is(x∗, y∗) = (−1, 5). Because L(x, y) is concave, we have found the optimal point.

y

1

2

3

4

5

6

7

x−2 −1 1 2 3 4 5 6 7

II

(−1, 5)

I

III

(3, 1)

Figure M14.8.2

4. (a) The Lagrangian is L = y − x2 + λy + µ(y − x + 2) − ν(y2 − x), which is stationary when(i) −2x − µ + ν = 0; (ii) 1 + λ + µ − 2νy = 0. Complementary slackness requires in addition,



(iii) λ ≥ 0 (λ = 0 if y > 0), (iv) µ ≥ 0 (µ = 0 if y − x + 2 > 0); (v) ν ≥ 0 (ν = 0 if y2 < x).From (ii), 2νy = 1 + λ + µ > 0, so y > 0. Then (iii) implies λ = 0, and 2νy = 1 + µ. From (i),

x = 12 (ν − µ). But x ≥ y2 > 0, so ν > µ ≥ 0, and from (v), y2 = x.Suppose µ > 0. Then y − x + 2 = y − y2 + 2 = 0 with roots y = −1 and y = 2. Only y = 2 is

feasible. Then x = y2 = 4. Because λ = 0, conditions (i) and (ii) become −µ+ν = 8 and µ−4ν = −1,so ν = −7/3, which contradicts ν ≥ 0, so (x, y) = (4, 2) is not a candidate. Therefore µ = 0 afterall. Thus x = 1

2ν = y2 and, by (ii), 1 = 2νy = 4y3. Hence y = 4−1/3, x = 4−2/3. This is the onlyremaining candidate. It is the solution with λ = 0, µ = 0, and ν = 1/2y = 4−1/6.

(b) We write the problem as max xey−x − 2ey subject to y ≤ 1 + 12x, x ≥ 0, y ≥ 0. The Lagrangian is

L = xey−x − 2ey − λ(y − 1 − x/2), so the first-order conditions (4) and (5) are:(i) L′

x = ey−x − xey−x + 12λ ≤ 0 (= 0 if x > 0); (ii) L′

y = xey−x − 2e − λ ≤ 0 (= 0 if y > 0);

(iii) λ ≥ 0 (λ = 0 if y < 1 + 12x).

From (i) we have x ≥ 1 + 12λex−y ≥ 1, so (iv) (x − 1)ey−x = 1

2λ. Suppose λ > 0. Then (iii)implies (v) y = 1 + 1

2x > 0. From (ii) and (iv) we then have xey−x = 2e + λ = ey−x + 12λ and so

λ = 2ey−x −4e = 2e(e− 12 x −2), by (v). But then λ > 0 implies that e− 1

2 x > 2, which contradicts x ≥ 0.When λ = 0, (iv) gives x = 1. If y > 0, then (ii) yields ey−1 = 2e, and so y − 1 = ln(2e) > 1

2x whenx = 1. Thus feasibility requires that y = 0, so we see that (x, y) = (1, 0) is the only point satisfying allthe conditions, with λ = 0.


2. (a) x = 2m

5p, y = 3m

5q(b) x = m

3p, y = 2m

3q(c) x = 3m

5p, y = 2m

5q

4. (a) Interpretation: Maximize the volume of a rectangular box with a given surface area A.(b) With the Lagrangian L = xyz − λ(2xy + 2xz + 2yz − A), the first-order conditions yieldyz − λ(2y + 2z) = 0, xz − λ(2x + 2z) = 0, xy − λ(2x + 2y) = 0. (The optimal solution obviously

requires all three variables to be positive.) It follows that λ = yz

y + z= xz

x + z= xy

x + y. The other

required equality is just the constraint divided by 2.(c) From yz/(y + z) = xz/(x + z) we get y(x + z) = x(y + z), or yz = xz, so x = y because z > 0.Similarly, from xz/(x + z) = xy/(x + y) we get y = z. Inserting y = z = x into the constraint andsolving for x yields x = √

A/6, and so the optimal solution is x = y = z = √A/6. The optimal box is

a cube whose six faces all have area A/6.

6. With the Lagrangian L = −(1/x)+ 4y + 4z−λ(x + y + z− 2)−µ(x2 + y2 + z2 − 3/2

), the first-order

conditions yield (i) 1/x2 − λ − 2µx = 0; (ii) 4 − λ − 2µy = 0; (iii) 4 − λ − 2µz = 0. Equations (ii)and (iii) give µy = µz, i.e. (iv) µ(y − z) = 0.

A. Suppose µ �= 0. Then (iv) implies y = z, which inserted into the constraints gives x = 2 − 2y andx2 + 2y2 = 3/2, so (2 − 2y)2 + 2y2 = 3/2, or 6y2 − 8y + 5/2 = 0. The solutions of this equationare y = 5/6 and y = 1/2. This gives the following solution candidates: P1 = (1/3, 5/6, 5/6) andP2 = (1, 1/2, 1/2). The function values are are 11/3 and 3 respectively.

B. Suppose µ = 0. From (ii) we have λ = 4, and then (i) gives 1/x2 = 4, so x = ±1/2. Inserting x = 1/2into the first constraint gives z = 3/2−y, and then the second constraint gives 2y2 −3y +1 = 0, with theroots y = 1 and y = 1/2. The new solution candidates are P3 = (1/2, 1, 1/2) and P4 = (1/2, 1/2, 1).The function values are both 4.



It remains to consider the case x = −1/2. But then the first constraint gives z = 5/2 − y, and thesecond constraint gives the equation 2y2 − 5y + 5 = 0, which has no real solutions.

By evaluating the objective function f at all the four candidate points, we find that f has its largestvalue at P3 and P4, with fmax = f (1/2, 1, 1/2) = f (1/2, 1/2, 1) = 4.

8. (a) The maximum value is 20 attained at (x, y, z) = (4, 0, 0). The minimum value is 7.5 attained at allpoints (x, y, z) satisfying x = −1 and y2 + z2 = 7.5. (With λ as the Lagrange multiplier, the first-orderconditions imply (i) 2x + 1 = 2λx; (ii) 2y = 4λy; (iii) 2z = 4λz. From (ii), y = 0 or λ = 1/2. From(iii), z = 0 or λ = 1/2. Suppose first that λ = 1/2. Then (i) implies x = −1 and 2(y2 + z2) = 15,which gives the minimum value. On the other hand, if λ �= 1/2, then (ii) and (iii) give y = z = 0. Thenx = ±4, and x = 4 gives the maximum. By the extreme value theorem, maximum and minimum valuesexist in (a) as well as in (b).)(b) The function f has a unique stationary point in the interior of S at (x, y, z) = (−1/2, 0, 0), withf (−1/2, 0, 0) = −1/4. We conclude that the maximum point of f in S is at (4, 0, 0) and the minimumis at (−1/2, 0, 0).

10. (a) The Lagrangian is L(x, y) = U(x, y) − λ[py − w(24 − x)]. The first-order conditions implypU ′

1 = wU ′2 = λwp, which immediately yields (∗∗). (b) Differentiating (∗) and (∗∗) w.r.t. w gives

py ′w = 24 − x − wx ′

w and p(U ′′11x

′w + U ′′

12y′w) = U ′

2 + w(U ′′21x

′w + U ′′

22y′w). Solving for x ′

w = ∂x/∂w

yields the given formula. (c) ∂x/∂w = −(4 ln 18 − 5)/(ln 18 + 1).

12. (a) See Fig. M14.R.12. Note that x ≥ ln(3/2). (b) With λ and µ as the Lagrange multipliers, the first-order conditions are: (i) −(2x + 1) + λe−x = 0; (ii) −y + λ − µ = 0; (iii) λ ≥ 0 (λ = 0 if e−x < y);(iv) µ ≥ 0 (µ = 0 if y < 2/3). From (i), λ = (2x + 1)ex ≥ 3/2, because of (a). From (ii), µ =λ − y ≥ 3/2 − 2/3 > 0, so y = 2/3 because of (iii). Solution: (x∗, y∗) = (ln(3/2), 2/3), withλ = 3[ln(3/2) + 1/2], µ = 3 ln(3/2) + 5/6. (The Lagrangian is concave so this is the solution.)

y

1

2

x1 2

y = 2/3

y = e−x

(x∗, y∗)

Figure M14.R.12


C H A P T E R 1 5 M A T R I X A N D V E C T O R A L G E B R A 67

Chapter 15 Matrix and Vector Algebra

Chapter 15 begins our treatment of matrix algebra by presenting matrices, vectors, and simple algebraic rulesfor working with them. Section 15.1 introduces a general notation for linear simultaneous equations, andmotivates them through a simple case of the Leontief input–output system.

Matrices are defined in Section 15.2 as rectangular arrays of numbers. (The idea that matrices representlinear transformations is, in our view, better postponed to a more advanced and abstract book on linearalgebra.) Particular examples are the coefficient matrix for a system of linear simultaneous equations, anda “spreadsheet” recording the value of retail sales for different commodities at different sales outlets. Thenequality, addition, and multiplication by a scalar are defined in obvious ways.

The rules for matrix multiplication are far from intuitively obvious to a student seeing them for the firsttime. Section 15.3 therefore takes some pains to explain how C = AB is defined for a 2 × 3 matrix A and a3×2 matrix B in order that the 2×2 matrix C should satisfy Cx = A(Bx) for all 2-vectors x. This, of course,builds on the rules for multiplying a coefficient matrix by a vector, which are also explained in Section 15.3.Thereafter, the rule for multiplying general m × n and n × p matrices is set out and illustrated. Attention isdrawn to the important fact that matrix multiplication is generally noncommutative—i.e., generally AB �= BA.The section concludes by reviewing how a general system of equations can be expressed concisely in matrixform.

Next, Section 15.4 turns to the associative law and (since multiplication is non-commutative) the twoversions of the distributive law for matrix multiplication. Then powers of square matrices are defined, as isthe identity matrix of any finite dimension. The section concludes by warning against algebraic errors suchas replacing AB by BA, or asserting that if AB = 0, then either A = 0 or B = 0 (or both).

Section 15.5 gives a brief discussion of transpose matrices, including general rules for transposition, andof symmetric matrices.

A powerful general method for solving linear systems of equations is Gaussian elimination, which isthe subject of Section 15.6. The “staircase” method is explained and a convenient notation for elementaryoperations is introduced.

Next, Section 15.7 introduces n-dimensional vectors as lists of n numbers. The rules of addition, sub-traction, and multiplication by a scalar are described. Linear combinations are defined.

Particularly for two-dimensional vectors, the fact that there is a simple geometric interpretation can helpstudents enormously. This is the topic of Section 15.8, which also discusses the parallelogram law of vectoraddition, and how to represent multiplication by a scalar. The section concludes with some remarks aboutn-dimensional geometry.

Calculating the inner (or scalar) product of two n-vectors is an important step on the way to findingthe product of two matrices. In economics, it is common to consider corresponding vectors of prices and ofquantities. Then the inner product of a price and quantity vector represents the value of a commodity bundle.This is used in Section 15.8 to motivate the concept. Rules for manipulation are presented. The length of avector is defined. The Cauchy–Schwarz inequality is stated. The section concludes with a brief discussion oforthogonality, a concept that is useful in econometrics.

Geometric ideas receive more attention in Section 15.9. A diagram is used to motivate the algebraicdefinition of a line in a space of three dimensions. This algebraic definition is then extended to n dimensions.Similarly, a diagram of a plane in three dimensions is used to motivate the algebraic definition, which is thenextended to n dimensions.


68 C H A P T E R 1 5 M A T R I X A N D V E C T O R A L G E B R A


15.1

2. Yes, the system is linear in a, b, c, and d .

4. 2x1 + 3x2 + 4x3 = 13x1 + 4x2 + 5x3 = 24x1 + 5x2 + 6x3 = 3

6. (a) The commodity bundle owned by individual j . (b) ai1+ai2+· · ·+ain is the total stock of commodity i.The first case is when i = 1. (c) p1a1j + p2a2j + · · · + pmamj

15.2

2. u = 3 and v = −2. (Equating the elements in row 1 and column 3 gives u = 3. Then, equating those inrow 2 and column 3 gives u − v = 5 and so v = −2. The other elements then need to be checked, butthis is obvious.)

4. A + B =(

1 0 42 4 16

), A − B =

(−1 2 −62 2 −2

), and 5A − 3B =

(−3 8 −2010 12 8

)

15.3

2. (i)

(−1 15−6 −13

)(ii) and (iii): AB = C(AB) =

(0 00 0

), since AB =

(0 00 0

).

4. (a)

(1 13 5

)(x1

x2

)=(

35

)(b)

⎛⎝ 1 2 11 −1 12 3 −1

⎞⎠⎛⎝ x1

x2

x3

⎞⎠ =⎛⎝ 4

51

⎞⎠(c)

(2 −3 11 1 −1

)⎛⎝ x1

x2

x3

⎞⎠ =(

00

)

6. T(Ts) =⎛⎝ 0.2875

0.22500.4875

⎞⎠15.4

2. (ax2 + by2 + cz2 + 2dxy + 2exz + 2fyz) (a 1 × 1 matrix)

4. Equality in (a) as well as in (b) if and only if AB = BA. ((A+B)(A−B) = A2−AB+BA−B2 �= A2−B2

unless AB = BA. The other case is similar.)

6. (a) Direct verification by matrix multiplication. (b) AA = (AB)A = A(BA) = AB = A, so A isidempotent. Then just interchange A and B to show that B is idempotent.(c) As the induction hypothesis, suppose that Ak = A, which is true for k = 1. Then Ak+1 = AkA =AA = A, which completes the proof by induction.



15.5

2. A′ =(

3 −12 5

), B′ =

(0 22 2

), (A + B)′ =

(3 14 7

), (αA)′ =

(−6 2−4 −10

),

AB =(

4 1010 8

), (AB)′ =

(4 10

10 8

)= B′A′, and A′B′ =

(−2 410 14

)Verifying the rules in (2) is now very easy.

4. Symmetry requires a2 − 1 = a + 1 and a2 + 4 = 4a. The second equation has the unique root a = 2,which also satisfies the first equation.

6. By the associative law and rule (15.5.2)(d) for transposition, (A1A2A3)′ = (A1(A2A3))

′ = (A2A3)′A′

1 =(A′

3A′2)A

′1 = A′

3A′2A′

1. In general, for any natural number n > 3, one has(A1A2 · · · An)

′((A1A2 · · · An−1)An)′ = A′

n(A1A2 · · · An−1)′. As the induction hypothesis, suppose the

result is true for n − 1. Then the last expression becomes A′nA′

n−1 · · · , A′2A′

1, so the result is true for n.

15.6

2. Gaussian elimination yields:⎛⎝ 1 1 −1 11 −1 2 21 2 a b

⎞⎠ −1 −1←←

∼⎛⎝ 1 1 −1 1

0 −2 3 10 1 a + 1 b − 1

⎞⎠ − 12

∼⎛⎝ 1 1 −1 1

0 1 −3/2 −1/20 1 a + 1 b − 1

⎞⎠ −1←

∼⎛⎝ 1 1 −1 1

0 1 −3/2 −1/20 0 a + 5/2 b − 1/2

⎞⎠For any z, the first two equations imply that y = − 1

2 + 32z and x = 1 − y + z = 3

2 − 12z. From the last

equation we see that for a �= − 52 , there is a unique solution with z = (b − 1

2 )/(a + 5

2 ). For a = − 52 ,

there are no solutions if b �= 12 , but z is arbitrary (one degree of freedom) if b = 1

2 .

4. (a) After moving the first row down to row number three, Gaussian elimination yields the matrix⎛⎝ 1 2 1 b2

0 1 −2 32b2 − 1

2b3

0 0 3 − 4a b1 + (2a − 32 )b2 + ( 1

2 − a)b3

⎞⎠Obviously, there is a unique solution iff a �= 3/4.

(b) Put a = 3/4 in part (a). Then the last row in the matrix in (a) becomes (0, 0, 0, b1 − 14b3). It follows

that if b1 �= 14b3 there is no solution. If b1 = 1

4b3 there are an infinite number of solutions. In fact,x = −2b2 + b3 − 5t , y = 3

2b2 − 12b3 + 2t , z = t , with t ∈ �.

15.7

2. a + b + c = (−1, 6, −4), a − 2b + 2c = (−3, 10, 2), 3a + 2b − 3c = (9, −6, 9),−a − b − c = −(a + b + c) = (1, −6, 4).

4. (a) xi = 0 for all i. (b) Nothing, because 0 · x = 0 for all x.

6. 4x − 2x = 7a + 8b − a, so 2x = 6a + 8b, and x = 3a + 4b.

8. The scalar product of the two vectors is x2+(x−1)x+3·3x = x2+x2−x+9x = 2x2+8x = 2x(x+4),which is 0 for x = 0 and x = −4.


70 C H A P T E R 1 5 M A T R I X A N D V E C T O R A L G E B R A

10. (a) The firm’s revenue is p·z. Its costs are p·x. (b) Profit = revenue – costs = p·z−p·x = p·(z−x) = p·y.If p · y < 0, the firm makes a loss equal to −p · y.

15.82. (a) λ = 0 gives x = (3, 1), λ = 1/4 gives x = (2, 5/4), λ = 1/2 gives x = (1, 3/2),

λ = 3/4 gives x = (0, 7/4), and λ = 1 gives x = (−1, 2). See Fig. M15.8.2.(b) As λ runs through [0, 1], the vector x will run through all points on the line segment S between a andb. In fact, according to the point–point formula, the line L through (3, 1) and (−1, 2) has the equationx2 = − 1

4x1 + 74 or x1 + 4x2 = 7. The line segment S is traced out by having x1 run through [3, −1] as x2

runs through [1, 2]. Now, (1−λ)a+λb = (3−4λ, 1+λ). Any point (x1, x2) on L satisfies x1 +4x2 = 7and equals (3 − 4λ, 1 + λ) for λ = 1

4 (3 − x1) = x2 − 1. Any point on the segment of this line betweena = (3, 1) and b = (−1, 2) equals (3 − 4λ, 1 + λ) for some λ ∈ [0, 1].

y

x

1

−1 1 2 3

ba

λ = 1λ = 3/4

λ = 1/2λ = 1/4

λ = 0

Figure M15.8.2

4. (a) x1(1, 2, 1) + x2(−3, 0, −2) = (x1 − 3x2, 2x1, x1 − 2x2) = (5, 4, 4) when x1 = 2 and x2 = −1.(b) x1 and x2 must satisfy x1(1, 2, 1) + x2(−3, 0, −2) = (−3, 6, 1). Then x1 − 3x2 = −3, 2x1 = 6, andx1 − 2x2 = 1. The first two equations yield x1 = 3 and x2 = 2; then the last equation is not satisfied.

6. The vectors are orthogonal iff their scalar product is 0, that is iff x2 −x − 8 − 2x +x = x2 − 2x − 8 = 0,which is the case for x = −2 and x = 4.

8. (||a|| + ||b||)2 − ||a + b||2 = ||a||2 + 2||a|| · ||b|| + ||b||2 − (a + b)(a + b) = 2(||a|| · ||b|| − a · b) ≥2(||a|| · ||b|| − |a · b|) ≥ 0 by the Cauchy–Schwarz inequality (2).

15.92. (a) Direct verification. (To show that a lies on L, put t = 0.) (b) The direction of L is given by the vector

(−1, 2, 1), and the equation of the orthogonal plane is (−1)(x1 − 2) + 2(x2 − (−1)) + 1 · (x3 − 3) = 0,or −x1 + 2x2 + x3 = −1. (c) We must have 3(−t + 2) + 5(2t − 1) − (t + 3) = 6, and so t = 4/3. Thepoint of intersection is then (2/3, 5/3, 13/3).

4. (a) Direct verification. (b) (x1, x2, x3) = (−2, 1, −1) + t (−1, 2, 3) = (−2 − t, 1 + 2t, −1 + 3t).


2. (a) A − B =(

3 −2−2 2

)(b) A + B − 2C =

(−3 −4−2 −8

)(c) AB =

(−2 42 −3

)(d) C(AB) =

(2 −16 −8

)(e) AD =

(2 2 20 2 3

)(f) DC is not defined.



4. (a)

(2 −55 8

)(x1

x2

)=(

35

)(b)

⎛⎜⎜⎝1 1 1 11 3 2 41 4 8 02 0 1 −1

⎞⎟⎟⎠⎛⎜⎜⎝

x

y

z

t

⎞⎟⎟⎠ =

⎛⎜⎜⎝a

b

c

d

⎞⎟⎟⎠(c)

⎛⎝ a 1 a + 11 2 13 4 7

⎞⎠⎛⎝ x

y

z

⎞⎠ =⎛⎝ b1

b2

b3

⎞⎠

6. (a) TS =⎛⎝ p3 + p2q 2p2q + 2pq2 pq2 + q3

12p3 + 1

2p2 + 12p2q p2q + pq + pq2 1

2pq2 + 12q2 + 1

2q3

p3 + p2q 2p2q + 2pq2 pq2 + q3

⎞⎠ = S because p + q = 1.

A similar argument shows that T2 = 12 T + 1

2 S. To prove the last equality, we do not have to consider theindividual elements: T3 = TT2 = T( 1

2 T + 12 S) = 1

2 T2 + 12 TS = 1

2 ( 12 T + 1

2 S) + 12 S = 1

4 T + 34 S.

(b) The appropriate formula is (∗) Tn = 21−nT + (1 − 21−n)S. This formula is correct for n = 1(and for n = 2, 3). Suppose (∗) is true for n = k. Then using the two first equalities in (a), Tk+1 =TTk = T(21−kT + (1 − 21−k)S) = 21−kT2 + (1 − 21−k)TS = 21−k( 1

2 T + 12 S) + (1 − 21−k)S =

2−kT + 2−kS + S − 2 · 2−kS = 2−kT + (1 − 2−k)S, which is formula (∗) for n = k + 1.

8. (a)

(1 4 12 2 8

) −2← ∼

(1 4 10 −6 6

)−1/6

∼(

1 4 10 1 −1

) ←−4

∼(

1 0 50 1 −1

)The solution is x1 = 5, x2 = −1.

(b)

⎛⎝ 2 2 −1 21 −3 1 03 4 −1 1

⎞⎠ ←← ∼

⎛⎝ 1 −3 1 02 2 −1 23 4 −1 1

⎞⎠ −2 −3←←

∼⎛⎝ 1 −3 1 0

0 8 −3 20 13 −4 1

⎞⎠ 1/8 ∼⎛⎝ 1 −3 1 0

0 1 −3/8 1/40 13 −4 1

⎞⎠ −13←

∼⎛⎝ 1 −3 1 0

0 1 −3/8 1/40 0 7/8 −9/4

⎞⎠8/7

∼⎛⎝ 1 −3 1 0

0 1 −3/8 1/40 0 1 −18/7

⎞⎠ ←3

∼⎛⎝ 1 0 −1/8 3/4

0 1 −3/8 1/40 0 1 −18/7

⎞⎠ ←←

3/8 1/8∼⎛⎝ 1 0 0 3/7

0 1 0 −5/70 0 1 −18/7

⎞⎠The solution is x1 = 3/7, x2 = −5/7, x3 = −18/7.

(c)

(1 3 4 05 1 1 0

) −5← ∼

(1 3 4 00 −14 −19 0

)−1/14

∼(

1 3 4 00 1 19/14 0

) ←−3

∼(

1 0 −1/14 00 1 19/14 0

)The solution is x1 = (1/14)x3, x2 = −(19/14)x3, where x3 is arbitrary. (One degree of freedom.)

10. ‖a‖ = √35, ‖b‖ = √

11, and ‖c‖ = √69. Also, |a · b| = |(−1) · 1 + 5 · 1 + 3 · (−3)| = | − 5| = 5, and√

35√

11 = √385 is obviously greater than 5, so the Cauchy–Schwarz inequality is satisfied.

12. If PQ − QP = P, then PQ = QP + P, and so P2Q = P(PQ) = P(QP + P) = (PQ)P + P2 =(QP+P)P+P2 = QP2+2P2. Thus, P2Q−QP2 = 2P2. Moreover, P3Q = P(P2Q) = P(QP2+2P2) =(PQ)P2 + 2P3 = (QP + P)P2 + 2P3 = QP3 + 3P3 Hence, P3Q − QP3 = 3P3.


72 C H A P T E R 1 6 D E T E R M I N A N T S A N D I N V E R S E M A T R I C E S

Chapter 16 Determinants and Inverse Matrices

In this chapter, Section 16.1 begins with determinants of order 2, whose definition is motivated by the form ofthe typical solution to two linear simultaneous equations in two unknowns. In fact, determinants are definedso that Cramer’s rule for solving such equations is valid. Then it is shown that the determinant of a 2×2 matrixis equal to ± the area of the parallelogram generated by its two columns, regarded as a pair of 2-vectors.

Next, Section 16.2 moves on to three equations in three unknowns. This gives rise to a complicatedformula, which is then broken down so that it appears as the expansion of the determinant by the cofactors ofits first row. It is claimed that the determinant of a 3 × 3 matrix equals ± the volume of the (3-dimensional)parallelepiped generated by its columns, regarded as three 3-vectors. Sarrus’s rule for evaluating 3 × 3determinants is then introduced.

Section 16.3 proposes a formula for n × n determinants that is essentially the sum of n! terms, each termbeing the product of n elements, of which one is chosen from each row and one from each column of thematrix. Then a sign rule is given to determine whether this product should have a + or − attached in formingthe overall sum. Of course, this formula is virtually useless except in a few special cases of “sparse” matriceswhere virtually all the n! terms collapse to 0.

In fact, Section 16.3 really helps clear the ground for the fundamental “Rules for Determinants” that areenunciated in Theorem 16.4.1. Particularly important is the rule saying that determinants are unchanged bythe elementary row (or column) operations of adding a multiple of one row (or column) to any other row (orcolumn). These rules are illustrated by showing how they apply to particular 2 × 2 matrices. There are alsotwo pages of discussion concerning how to prove most parts of Theorem 16.4.1. We envisage, however, thatmost students will prefer to take on trust the key results set out in the theorem.

The topic of Section 16.5 is expansion by cofactors in general—not just the expansion of 3×3 determinantsby the first row that was considered in Section 16.2. Also important is the fact that expansion by “alien”cofactors always gives zero, because it is like finding the determinant of a matrix in which one row or columnhas been repeated. The section concludes with a proof that the rule for expanding cofactors is valid, thoughonce again, this is a result that we expect most students will prefer to take on trust. By the end of this sectionstudents should have learned how to find the determinant of a matrix in any one of several equivalent ways—expansion by cofactors, the full expansion of n! terms, or by elementary row or column operations designedto produce a “sparse” matrix—even one that is upper or lower triangular, and so with determinant equal tothe product of its diagonal terms.

Having completed the discussion of determinants per se, Section 16.6 moves on to the inverse of a squarematrix. After the definition, it is claimed that an inverse exists iff the matrix is non-singular in the sense ofhaving a non-zero determinant. It is shown that, if an inverse exists, it must be unique. The general formula forthe inverse of a 2 × 2 matrix is given. Then Theorem 16.6.1 sets out some key properties of inverse matrices.These include the results that the inverse of the inverse is the original matrix, that the inverse of a matrixproduct is the reverse product of the inverses, and that the inverse of the transpose is equal to the transposeof the inverse. Finally, Theorem 16.6.2 states how to solve linear equations by matrix inversion, in case therelevant matrix is non-singular.

An explicit formula for the inverse of a matrix is the main topic of Section 16.7. Indeed, based on therules for expansion by cofactors (including alien cofactors), it is shown that one can find the adjoint matrix,defined as the transpose of the matrix of cofactors, then divide the adjoint by the (non-zero) determinant ofthe (non-singular) matrix. This is the “general formula for the inverse” set out in Theorem 16.7.1. The sectionconcludes by outlining how to find the inverse matrix by the generally more practical procedure of elementaryrow operations.


C H A P T E R 1 6 D E T E R M I N A N T S A N D I N V E R S E M A T R I C E S 73

Section 16.8 explains how a general linear system of n equations in n unknowns with a non-singularcoefficient matrix can be solved by means of Cramer’s rule (Theorem 16.8.1). This result is of theoreticalinterest, but for systems with more than 4 unknowns, the calculation of the appropriate determinants usuallybecomes computationally impractical. Finally, whereas a linear system of n equations in n unknowns isusually soluble only when its coefficient matrix is non-singular, a system of homogeneous equations (withzero right-hand sides) has a non-trivial solution iff the coefficient matrix is singular. This is Theorem 16.8.2.

To conclude the chapter, Section 16.9 gives a brief introduction to the Leontief model.


16.1

2. See Fig. M16.1.2. The shaded area is 18 =∣∣∣∣ 3 0

2 6

∣∣∣∣.

(3, 0)

(2, 6)

Figure M16.1.2

4. The matrix product is AB =(

a11b11 + a12b21 a11b12 + a12b22

a21b11 + a22b21 a21b12 + a22b22

), implying that

|AB| = (a11b11 + a12b21)(a21b12 + a22b22) − (a11b12 + a12b22)(a21b11 + a22b21). On the other hand,|A||B| = (a11a22 − a12a21)(b11b22 − b12b21). A tedious process of expanding each expression, thencancelling four terms in the expression of |A||B|, reveals that the two expressions are equal.

6. We write the system asY − C = I0 + G0

−bY + C = a

Then Cramer’s rule yields

Y =

∣∣∣∣ I0 + G0 −1a 1

∣∣∣∣∣∣∣∣ 1 −1−b 1

∣∣∣∣ = a + I0 + G0

1 − b, C =

∣∣∣∣ 1 I0 + G0

−b a

∣∣∣∣∣∣∣∣ 1 −1−b 1

∣∣∣∣ = a + b(I0 + G0)

1 − b

The expression for Y is most easily found by substituting the second equation into the first, and thensolving for Y . Then use C = a + bY to find C.

16.2

2. AB =⎛⎝−1 −1 −1

7 13 135 9 10

⎞⎠, |A| = −2, |B| = 3, |AB| = |A| · |B| = −6



4. By Sarrus’ rule the determinant is (1 + a)(1 + b)(1 + c) + 1 + 1 − (1 + b) − (1 + a) − (1 + c), whichreduces to the given expression.

6. (a) Substituting T = d + tY into the expression for C gives C = a − bd + b(1 − t)Y . Substituting forC in the expression for Y then yields Y = a + b(Y − d − tY ) + A0. Then solve for Y , T , and C in turnto derive the answers given in (b) below.

(b) We write the system as

⎛⎝ 1 −1 0−b 1 b

−t 0 1

⎞⎠⎛⎝ Y

C

T

⎞⎠ =⎛⎝A0

a

d

⎞⎠. Then Cramer’s rule yields

Y =

∣∣∣∣∣∣A0 −1 0a 1 b

d 0 1

∣∣∣∣∣∣∣∣∣∣∣∣1 −1 0

−b 1 b

−t 0 1

∣∣∣∣∣∣= a − bd + A0

1 − b(1 − t), C =

∣∣∣∣∣∣1 A0 0

−b a b

−t d 1

∣∣∣∣∣∣∣∣∣∣∣∣1 −1 0

−b 1 b

−t 0 1

∣∣∣∣∣∣= a − bd + A0b(1 − t)

1 − b(1 − t)

T =

∣∣∣∣∣∣1 −1 A0

−b 1 a

−t 0 d

∣∣∣∣∣∣∣∣∣∣∣∣1 −1 0

−b 1 b

−t 0 1

∣∣∣∣∣∣= t (a + A0) + (1 − b)d

1 − b(1 − t)

(This problem is meant to train you in using Cramer’s rule. Note how systematic elimination is muchmore efficient.)

16.3

2. +a12a23a35a41a54 (Four lines between pairs of boxed elements rise as one goes to the right.)

4. With A =

⎛⎜⎜⎝a11 a12 . . . a1n

0 a22 . . . a2n...

.... . .

...

0 0 . . . ann

⎞⎟⎟⎠ and B =

⎛⎜⎜⎝b11 b12 . . . b1n

0 b22 . . . b2n...

.... . .

...

0 0 . . . bnn

⎞⎟⎟⎠,

the product AB is easily seen to be upper triangular, with the elements a11b11, a22b22, . . . , annbnn on themain diagonal. The determinant |AB| is, according to (4), the product of these n numbers. On the otherhand, |A| = a11a22 · · · ann, and |B| = b11b22 · · · bnn, so the required equality follows immediately.

16.4

2. A′ =⎛⎝ 2 1 1

1 0 23 1 5

⎞⎠, |A| = |A′| = −2

4. |AB| = |A||B| = −12, 3|A| = 9, | − 2B| = (−2)3(−4) = 32, |4A| = 43|A| = 43 · 3 = 192,

|A|+|B| = −1, whereas |A+B| is not determined. (As an example showing this, let A1 =⎛⎝ 1 0 0

0 1 00 0 3

⎞⎠© Knut Sydsæter and Peter Hammond 2006


and B1 =⎛⎝ 1 0 0

0 1 00 0 −4

⎞⎠. Then |A1| = 3, |B1| = −4, and |A1 + B1| = −4. On the other hand, if

A2 =⎛⎝ 3 0 0

0 1 00 0 1

⎞⎠ and B2 =⎛⎝ 2 0 0

0 −2 00 0 1

⎞⎠, then |A2| = 3, |B2| = −4, and |A2 + B2| = −10.)

6. (a) The first and the second columns are proportional, so the determinant is 0 by part E of Theorem 16.4.1.(b) Add the second column to the third. This makes the first and third columns proportional.(c) The term x − y is a common factor for each entry in the first row. If x �= y, the first two rows areproportional. If x = y, the first row has all elements 0. In either case the determinant is 0.

8. By Sarrus’s rule, for example, |Aa| = a(a2 + 1)+ 4 + 4 − 4(a2 + 1)− a − 4 = a2(a − 4), so |A1| = −3and |A6

1| = |A1|6 = (−3)6 = 729.

10. (a) Because A2 = In it follows from part G of Theorem 16.4.1 that |A|2 = |In| = 1, and so |A| = ±1.(b) Direct verification by matrix multiplication.(c) (In − A)(In + A) = I2

n + A − A − A2 = In − A2 = 0 ⇐⇒ A2 = In.

12. Add each of the last n − 1 rows to the first row. Each element in the first row then becomes na + b. So

Dn = (na + b)

∣∣∣∣∣∣∣∣1 1 . . . 1a a + b . . . a...

.... . .

...

a a . . . a + b

∣∣∣∣∣∣∣∣Next, add the first row multiplied by −a to all the other n − 1 rows. The result is an upper triangularmatrix whose diagonal elements are 1, b, b, ..., b, with product equal to bn−1. The conclusion followseasily.

16.52. (a) −abc (b) abcd (c) 6 · 4 · 3 · 5 · 1 = 360

16.62. Multiply the two matrices to get I3.

4. (a)

(x

y

)=(

2 −33 −4

)−1( 35

)=(−4 3

−3 2

)(35

)=(

31

)(b)

(x

y

)=(−4 3

−3 2

)(811

)=(

1−2

)(c)

(x

y

)=(

00

)

6. (a) |A| = 1, A2 =⎛⎝ 0 1 1

1 1 21 1 1

⎞⎠, A3 =⎛⎝ 1 1 2

2 2 31 2 2

⎞⎠. Direct verification yields A3 − 2A2 + A − I = 0.

(b) The equality shown in (a) is equivalent to A(A − I)2 = I, so A−1 = (A − I)2.

(c) Choose P = (A − I)−1 =⎛⎝ 0 0 1

1 0 10 1 0

⎞⎠, so that A = [(A − I)2]−1 = P2. The matrix −P also works.



8. (a) A2 = (PDP−1)(PDP−1) = PD(P−1P)DP−1 = PDIDP−1 = P(D)2P−1.(b) Suppose the formula is valid for m = k. Then Ak+1 = AAk = PDP−1(PDkP−1) = PD(P−1P)DkP−1

= PDIDkP−1 = PDDkP−1 = PDk+1P−1.

10. (a) If C2 + C = I, then C(C + I) = I, and so C−1 = C + I = I + C.(b) Because C2 = I − C, it follows that C3 = C2C = (I − C)C = C − C2 = C − (I − C) = −I + 2C.Moreover, C4 = C3C = (−I + 2C)C = −C + 2C2 = −C + 2(I − C) = 2I − 3C.

12. (a) AT = 112

⎛⎝ s + 17 4t − 16 02s + 10 5t − 8 03s + 15 4t − 16 12

⎞⎠ = I3, and so T = A−1, for s = −5, t = 4.

(b) BX = 2X + C ⇐⇒ BX − 2X = C ⇐⇒ (B − 2I)X = C. But it is easy to see that B − 2I = A.So BX = 2X + C ⇐⇒ AX = C ⇐⇒ X = A−1C = TC, when s = −5 and t = 4. Hence,

X = 112

⎛⎝−5 4 37 −8 31 4 −3

⎞⎠⎛⎝ 2 3 0 11 0 3 10 5 −4 1

⎞⎠ =⎛⎝−1/2 0 0 1/6

1/2 3 −3 1/61/2 −1 2 1/6

⎞⎠16.7

2. The determinant is 72. The cofactors are

C11 =∣∣∣∣ 0 3

1 −1

∣∣∣∣ = −3,

C21 = −∣∣∣∣ 3 2

1 −1

∣∣∣∣ = 5,

C31 =∣∣∣∣ 3 2

0 3

∣∣∣∣ = 9,

C12 = −∣∣∣∣ 6 3

4 −1

∣∣∣∣ = 18,

C22 =∣∣∣∣ −2 2

4 −1

∣∣∣∣ = −6,

C32 = −∣∣∣∣ −2 2

6 3

∣∣∣∣ = 18,

C13 =∣∣∣∣ 6 0

4 1

∣∣∣∣ = 6

C23 = −∣∣∣∣ −2 3

4 1

∣∣∣∣ = 14

C33 =∣∣∣∣ −2 3

6 0

∣∣∣∣ = −18

so the inverse is1

|A|

⎛⎝C11 C21 C31

C12 C22 C32

C13 C23 C33

⎞⎠ = 1

72

⎛⎝−3 5 918 −6 18

6 14 −18

⎞⎠.

4. Let B denote the n × p matrix whose ith column has the elements b1i , b2i , . . . , bni . The p systems ofn equations in n unknowns can be expressed as AX = B, where A is n × n and X is n × p. Followingthe method illustrated in Example 2, exactly the same row operations that transform the n × 2n matrix(A : I) into (I : A−1) will also transform the n × (n + p) matrix (A : B) into (I : B∗), whereB∗ is the matrix with elements b∗

ij . (In fact, because these row operations are together equivalent topremultiplication by A−1, it must be true that B∗ = A−1B.) When k = r , the solution to the system isx1 = b∗

1r , x2 = b∗2r , . . . , xn = b∗

nr .

16.8

2. The determinant of the system is equal to −10, so the solution is unique. The determinants in (2) are here

D1 =∣∣∣∣∣∣b1 1 0b2 −1 2b3 3 −1

∣∣∣∣∣∣ , D2 =∣∣∣∣∣∣

3 b1 01 b2 22 b3 −1

∣∣∣∣∣∣ , D3 =∣∣∣∣∣∣

3 1 b1

1 −1 b2

2 3 b3

∣∣∣∣∣∣© Knut Sydsæter and Peter Hammond 2006


Expanding each of these determinants by the column (b1, b2, b3), we find that D1 = −5b1 + b2 + 2b3,D2 = 5b1 −3b2 −6b3, D3 = 5b1 −7b2 −4b3. Hence, x1 = 1

2b1 − 110b2 − 1

5b3, x2 = − 12b1 + 3

10b2 + 35b3,

x3 = − 12b1 + 7

10b2 + 25b3.

16.92. (a) No sector delivers to itself. (b) The total amount of good i needed to produce one unit of each good.

(c) This column vector gives the number of units of each good which are needed to produce one unit ofgood j . (d) No meaningful economic interpretation. (The goods are usually measured in different units,so it is meaningless to add them together.)

4. The Leontief system for this three-sector model is as follows:

0.9x1 − 0.2x2 − 0.1x3 = 85−0.3x1 + 0.8x2 − 0.2x3 = 95−0.2x1 − 0.2x2 + 0.9x3 = 20

which does have the claimed solution.

6. The quantity vector x0 must satisfy (∗) (In − A)x0 = b and the price vector p′0 must satisfy (∗∗)

p′0(In−A) = v′. Multiplying (∗∗) from the right by x0 yields v′x0 = (p′

0(In−A))x0 = p′0((In−A)x0) =

p′0b.


2. (a) −4 (b) 1. (Suggestion: Subtract row 1 from rows 2 and 3. Then subtract twice row 2 from row 3.The resulting determinant has only one non-zero term in its third row.) (c) 1. (Suggestion: Use exactlythe same row operations as in (b).)

4. (a) |At | = t + 1, so At has an inverse iff t �= −1. (b) Multiplying the given equation from the right by

A1 yields BA1 + X = I3. Hence X = I3 − BA1 =⎛⎝ 0 0 −1

0 0 −1−2 −1 0

⎞⎠.

6. The determinant of the coefficient matrix is

∣∣∣∣∣∣−2 4 −t

−3 1 t

t − 2 −7 4

∣∣∣∣∣∣ = 5t2 − 45t + 40 = 5(t − 1)(t − 8).

So by Cramer’s rule, there is a unique solution iff t �= 1 and t �= 8.

8. (a) (In + aU)(In + bU) = I2n + bU + aU + abU2 = In + (a + b +nab)U, because U2 = nU, as is easily

verified. (b) Note first that

⎛⎝ 4 3 33 4 33 3 4

⎞⎠ = I3 + 3U. Moreover, (a) implies that

(I3 + 3U)(I3 + bU) = I3 + (3 + 10b)U = I3 for b = −3/10. Thus the inverse of

⎛⎝ 4 3 33 4 33 3 4

⎞⎠ is

I3 − (3/10)U = 1

10

⎛⎝ 7 −3 −3−3 7 −3−3 −3 7

⎞⎠.

10. (a) |A| = −2. A2 − 2I2 =(

11 −618 −10

)= A, so A2 + cA = 2I2 if c = −1.

(b) If B2 = A, then |B|2 = |A| = −2, which is impossible.



12. (a) For a �= 1 and a �= 2, there is a unique solution. If a = 1, there is no solution. If a = 2, there areinfinitely many solutions. (These are a special case of the results in (b).)(b) Using Gaussian elimination (where we start by interchanging rows 1 and 3), we obtain the matrix⎛⎝ 1 0 −3 b3

0 1 a2 + 6 b2 − 2b3

0 0 −a2 + 3a − 2 b1 − b2 + (2 − a)b3

⎞⎠The last row corresponds to the equation (−a2 + 3a − 2)z = b1 − b2 + (2 − a)b3. If −a2 + 3a − 2 =−(a − 1)(a − 2) �= 0, i.e. if a �= 1 and a �= 2, then the system has a unique solution. When a = 1 andb1 − b2 + b3 = 0, or when a = 2 and b1 = b2, there are infinitely many solutions. Otherwise there is nosolution.

14. Using Gaussian elimination, one can show that there is a unique solution when a �= 0 and b �= 2. Thesolution is:

x = 2b − 2a + 3

a(b − 2), y = 2a + b − 9

b − 2, z = 2ab − 2a − 2b − 3

a(b − 2), u = 7 − 2a

b − 2


C H A P T E R 1 7 L I N E A R P R O G R A M M I N G 79

Chapter 17 Linear Programming

Chapter 17 presents some of the main ideas behind linear programming (LP). But it does not cover the simplexmethod. This we regard as an important computational technique that is suitable for a specialist course, butnot really essential knowledge for most economists.

Section 17.1 begins by considering problems with only two choice variables, which can therefore besolved graphically. Example 1 is a detailed discussion of a simple introductory “baker’s problem”; Example 2moves rather faster. Then, after a brief discussion of when graphical techniques will work, the latter part ofthe section presents the general LP problem. It also gives an n-dimensional geometric interpretation of suchproblems, enabling the basic idea of the simplex method to be visualized.

Duality theory is introduced in Section 17.2. Example 1 revisits the baker’s problem and explicitlycalculates the marginal profit from relaxing each of the three main constraints. The resulting numbers (whichwill be interpreted as shadow prices later) are then used to verify algebraically the solution found by geometrictechniques. Then it is shown that, in this same example, these shadow prices solve the dual problem. This isset up as a problem facing somebody who wants to pay as little as possible to take over the resources availableto the baker, while offering just enough to ensure that the baker is compensated sufficiently for these resources.The resulting prices then enable the baker to earn just as much profit as by continuing in business. After theexample, it is shown how to set up the dual of a general LP problem. Finally, the concise matrix formulationsof the primal and dual problems are presented.

Section 17.3 begins with a further discussion of the baker’s problem. This time, the point is made thatany price vector that is feasible for the dual problem must give a value for the resources that is not less than theprofit earned from any activity vector that is feasible for the primal. Indeed, this is stated as Theorem 17.3.1,and proved in general linear programs by a very easy argument (which does, however, involve recognizinghow the same double sum can be written in two different ways). Moreover, Theorem 17.3.2 shows that ifthe value of primal and dual are made equal, then both problems have been solved. This has an even easierproof. The section concludes with the Duality Theorem 17.3.3, claiming: (i) that if the primal has an optimalsolution, then so does the dual; (ii) the values of both primal and dual must be equal when both are solved;and (iii) unless there is an upper bound in the value of the primal, the dual is infeasible. Proving (i) and (ii)involves relatively advanced arguments (or at least an understanding of the simplex method) so the proofs areomitted. An easy proof of (iii) is provided, however.

Section 17.4 provides a standard economic interpretation of any LP problem in terms of activity vectorsand scarce resources. It also interprets shadow prices as those ensuring that the “shadow profit” on any activityis non-positive, and interprets the dual problem in rather the same way as was done in Section 17.2. Finally,it is shown how shadow prices can be used to estimate changes in the optimized value of the primal fromchanges in the resources.

The last topic of the chapter is complementary slackness, the subject of Section 17.5. The baker’s problemis used to illustrate how positive activity levels are associated with binding constraints in the dual, and positiveshadow prices with binding constraints in the primal. Then, for the case of a general primal with two variablesand three constraints, the part of the duality theorem claiming equality of the values of primal and dual isinvoked. It shows that the pattern observed in the baker’s problem always holds. This complementary slacknessproperty is then stated and proved as Theorem 17.5.1. (The fairly long proof is in small print, especially asit uses vector and matrix notation at the beginning. But it is not hard.) The standard interpretations of thecomplementary slackness conditions are then presented. Example 1 is provided to show how, given thesolution to the dual problem with two variables (which can be found graphically, of course), complementaryslackness can be used to identify the set of binding constraints in the primal, and also the activity levels which


80 C H A P T E R 1 7 L I N E A R P R O G R A M M I N G

must be zero. Then, of course, the primal can be solved from simultaneous equations. This is an importanttechnique for solving (and setting) examination questions! Next, it is shown that the Kuhn–Tucker conditionsdiscussed in Chapter 14 can be applied to LP problems, and imply complementary slackness. This is importantin linking the results of Chapters 17 and 14. Finally, there is a brief discussion of what happens when someconstraints in the primal must hold with equality. Then the signs of the corresponding dual variables are nolonger constrained to be nonnegative.

17.12. (a) A graph shows that the solution is at the intersection of the lines −2x1 + 3x2 = 6 and x1 + x2 = 5.

Hence max = 98/5 for (x1, x2) = (9/5, 16/5).

(b) The solution satisfies 2x1 + 3x2 = 13 and x1 + x2 = 6. Hence max = 49 for (x1, x2) = (5, 1)

(c) The solution satisfies x1 − 3x2 = 0 and x2 = 2. Hence max = −10/3 for (x1, x2) = (2, 2/3)

4. (a) No maximum exists. Consider Fig. M17.1.4. By increasing c, the dashed level curve x1 + x2 = c

moves to the north-east and so this function can take arbitrarily large values.(b) Yes, the maximum is at P = (1, 0). The level curves are the same as in (a), but the direction ofincrease is reversed. x2

−1

1

2

3

4

x1−1 1 2 3 4 5 6

−x1 + 3x2 = 3

−x1 + x2 = −1

x1 + x2 = c

Figure M17.1.4

6. The LP problem is: max 700x + 1000y subject to

⎧⎨⎩3x + 5y ≤ 3900x + 3y ≤ 2100

2x + 2y ≤ 2200x ≥ 0 , y ≥ 0

A figure showing the admissible set and an appropriate level line for the criterion function will show thatthe solution is at the point where the two lines 3x + 5y = 3900 and 2x + 2y = 2200 intersect. Solvingthese equations yields x = 800 and y = 300. The firm should produce 800 sets of type A and 300 oftype B.

17.2

2. min 8u1 + 13u2 + 6u3 subject to

{u1 + 2u2 + u3 ≥ 8

2u1 + 3u2 + u3 ≥ 9u1 ≥ 0, u2 ≥ 0, u3 ≥ 0

17.3

2. max 300x1 + 500x2 subject to

{10x1 + 25x2 ≤ 10 00020x1 + 25x2 ≤ 8 000

x1 ≥ 0 , x2 ≥ 0

The solution can be found graphically. It is x∗1 = 0, x∗

2 = 320, and the value of the criterion function is160 000, the same value found in Example 17.1.2 for the optimal value of the primal criterion function.


C H A P T E R 1 7 L I N E A R P R O G R A M M I N G 81

17.5

2. (a) Figure M17.5.2 shows the relevant portion of the feasible set with the different constraints numbered,as well as two dashed level curves for the criterion function Z = y1 + 2y2. We see that the minimum isattained at the point (y∗

1 , y∗2 ) = (3, 2).

y2

5

10

y15 10 15

(3)

y1 + 2y2 = Z0

(1)(3, 2)

(2)

(4)

Figure M17.5.2

(b) The dual: max 15x1 + 5x2 − 5x3 − 20x4 s.t.x1 + x2 − x3 + x4 ≤ 1

6x1 + x2 + x3 − 2x4 ≤ 2, xj ≥ 0, j = 1, . . . , 4.

Because y∗1 and y∗

2 are both positive, both constraints are satisfied with equality. Furthermore, the two lastconstraints in the primal problem are satisfied with strict inequality at the optimum. Hence x∗

3 = x∗4 = 0.

The system therefore reduces to the system x∗1 + x∗

2 = 1, 6x∗1 + x∗

2 = 2, so x∗1 = 1/5 and x∗

2 = 4/5.The maximum is thus at (x∗

1 , x∗2 , x∗

3 , x∗4 ) = (1/5, 4/5, 0, 0).

(c) If the constraint is changed to y1 + 6y2 ≥ 15.1, the solution to the primal is still at the intersection ofthe lines (1) and (2) in Fig. M17.5.2, but with (1) shifted up slightly. The solution to the dual is completelyunchanged. The optimal values in both problems increase by (15.1 − 15) · x∗

1 = 0.02.

4. (a) For x3 = 0, the solution is x1 = x2 = 1/3. For x3 = 3, the solution is x1 = 1 and x2 = 2.(b) Let zmax denote the maximum value of the objective function. If 0 ≤ x3 ≤ 7/3, then zmax(x3) =2x3 + 5/3 for x1 = 1/3 and x2 = x3 + 1/3. If 7/3 < x3 ≤ 5, then zmax(x3) = x3 + 4 for x1 = x3 − 2and x2 = 5 − x3. If x3 > 5, then zmax(x3) = 9 for x1 = 3 and x2 = 0. Because zmax(x3) is increasing,the maximum is 9 for x3 ≥ 5.(c) The solution to the original problem is x1 = 3 and x2 = 0, with x3 as an arbitrary number ≥ 5.


2. (a) max −x1 + x2 s.t.

⎧⎪⎪⎨⎪⎪⎩−x1 + 2x2 ≤ 16

x1 − 2x2 ≤ 6−2x1 − x2 ≤ −8

−4x1 − 5x2 ≤ −15

x1 ≥ 0, x2 ≥ 0

The maximum value is 8 at (x1, x2) = (0, 8).(b) (y1, y2, y3, y4) = ( 1

2 (b + 1), 0, b, 0) for any b satisfying 0 ≤ b ≤ 1/5.(c) The maximand for the dual becomes kx1 + x2. The solution is unchanged provided that k ≤ −1/2.


82

TEST I ( Elementary Algebra)

A certain familiarity with elementary algebra is an essential prerequisite for reading the textbook (and forunderstanding most modern economics texts). This test is designed for students and instructors to discoverwhether the students have the proper background. (In a number of countries, many beginning economicsstudents’ background in elementary algebra appears to have become much weaker during the last few years.In fact, lecturers using this test (or similar ones) have been shocked by the results, and have had to readjusttheir courses.)

At the head of each problem, immediately after the number, the relevant sections of the introductorychapters in the book are given in parentheses, followed in square brackets by the number of points for acorrect answer to each separate part of the problem. In a 20–30 minute test, any student who scores less than50 (out of 100) has serious problems with elementary algebra. Such students definitely need to review therelevant section of Chapters 1 and 2, or consult other elementary material.

The correct answers are given on a separate page following the test.

1 (1.2) [Points: (a) 2; (b) 2; (c) 3; (d) 3] Calculate/simplify:

(a)73 · 72

74(b) (5.5 − 3.5)3 (c)

(−2

5

)(−2

5

)(−2

5

)(d)

219 − 217

219 + 217

2 (1.2–1.4) [Points: (a) 2; (b) 2; (c) 2; (d) 4]

(a) If 2x2y = 5, then 4x4y2 =? (b) 11 % of 3500 is? (c)√

132 − 122 =?

(d) Rationalize the denominator:

√3 + √

2√3 − √

2(i.e., find a new fraction that is equal but has no square root

in the denominator).

3 (1.3) [Points: (a) 2; (b) 2; (c) 2] Expand:

(a) (x + 2y)2 (b) (2x − 3y)2 (c) (a + b)(a − b)

4 (1.3) [Points: (a) 2; (b) 3; (c) 3; (d) 4] Expand and simplify:

(a) 5a − (3a + 2b) − 2(a − 3b) (b) (x + 2)2 + (x − 2)2 − 2(x + 2)(x − 2)

(c) (1 − x)2(1 + x)2 (d) (2 − a)3

5 (1.2) [Points: 4] If the GNP of a certain country in 2000 was 8 billion dollars, write down an expressionfor the GNP 6 years later if it increases by 5% each year.

6 (1.3) [Points: (a) 3; (b) 3; (c) 4] Factorize:

(a) 5a2b + 15ab2 (b) 9 − z2 (c) p3q − 4p2q2 + 4pq3


83

7 (1.4) [Points: (a) 2; (b) 2; (c) 2] Expand and simplify to a single fraction:

(a)1

2− 1

3(b)

6a

5− a

10+ 3a

20(c)

1

2− 1

3

1

4− 1

6

8 (1.5) [Points: (a) 2; (b) 2; (c) 2; (d) 2] Calculate/simplify:

(a) 251/2 (b) (x1/2y−1/4)4 (c) 3√

27a6 (d) p1/5(p4/5 − p−1/5)

9 (2.1) [Points: (a) 2; (b) 2; (c) 2] Solve the following equations for the unknown x:

(a)3

5x = −6 (b)

1

x − 1= 3

2x + 3(c)

√3 − x = 2

10 (1.6) [Points: (a) 2; (b) 3; (c) 3] Solve the following inequalities:

(a) −3x + 2 < 5 (b)x − 1

x + 3≤ 0 (c) x3 < x

11 (2.3) [Points: (a) 3; (b) 3; (c) 3] Solve the following equations:

(a) 3x − 9x2 = 0 (b) x2 − 2x − 15 = 0 (c) 2P 2 = 2 − 3P

12 (2.4) [Points: (a) 3; (b) 4; (c) 4] Solve the following systems of equations:

(a)2x − y = 5

x + 2y = 5(b)

1.5p − 0.5q = 14

2.5p + 1.5q = 28(c)

3

p+ 3

q= 3

3

p− 1

q= 7


84

TEST II (Elementary Mathematics)

Students who now enter university or college courses in economics tend to have a wide range of math-ematical backgrounds and aptitudes. At the low end, they may have no more than a shaky command ofelementary algebra. Or, at the high end, they may already have a ready facility with calculus, thoughoften it is some years since economics students took their last formal mathematics course. Experiencesuggests therefore that right from the start of the course, it is very important that the instructor, as wellas each individual student, should get some impression of what the student knows well, what is vaguelyfamiliar, and what seems to be more or less forgotten or perhaps never learned at all.

The present test is meant to test the students’ actual knowledge in some elementary mathematicaltopics of interest to economists. The level is nevertheless more advanced than for Test I and the topicscovered are discussed in the earlier chapters of the main text. The maximum total score is 100.

1 [Points: (a) 2; (b) 2; (c) 2; (d) 2]

(a) 25 + 25 = 2x �⇒ x = (b) 3−15 + 3−15 + 3−15 = 3y �⇒ y =

(c)√

132 − 52 = (d)226 − 223

226 + 223= z

9�⇒ z =

2 [Points: (a) 1+1; (b) 2+2]

(a) Find the slopes of the following straight lines:

(i) y = − 32x + 4 (ii) 6x − 3y = 5

(b) Find the equation of the straight line that:

(i) passes through (−2, 3) and has slope −2.(ii) passes through both (a, 0) and (0, b).

3 [Points: 5] Fill in the following table and sketch the graph of y = −x2 + 2x + 4.

x −2 −1 0 1 2 3 4

y = −x2 + 2x + 4

4 [Points: 4+4] Determine maximum/minimum points for:

(a) y = x2 − 4x + 8 (b) y = −2x2 + 16x − 14

5 [Points: 4+4] Perform the following polynomial divisions:

(a) (2x3 − 3x + 10) ÷ (x + 2) (b) (x4 + x) ÷ (x2 − 1)

6 [Points: 5] f (x) = 43x3 − 1

5x5. For what values of x is f ′(x) = 0?

7 [Points: 3+3+3+3] Sketch the graph of a function f in each of the following cases:(a) f ′(x) > 0 and f ′′(x) > 0, (b) f ′(x) > 0 and f ′′(x) < 0(c) f ′(x) < 0 and f ′′(x) > 0, (d) f ′(x) < 0 and f ′′(x) < 0


85

8 [Points: 3+3]

(a) The cost in dollars of extracting T tons of a mineral ore is given by C = f (T ). Give an economicinterpretation of the statement that f ′(1000) = 50.

(b) A consumer wants to buy a certain item at the lowest possible price. Let P(t) denote the lowestprice found after searching the market for t hours. What are the likely signs of P ′(t) and P ′′(t)?

9 [Points: 2+2+2+2] Write down the general rules for differentiating the following:

(a) y = f (x) + g(x) �⇒ y ′ = (b) y = f (x)g(x) �⇒ y ′ =(c) y = f (x)/g(x) �⇒ y ′ = (d) y = f (g(x)) �⇒ y ′ =

10 [Points: 2+2+2+2+2+2+2+2] Differentiate the following functions:

(a) y = x2 (b) y = x5/5 (c) y = x

x + 1(d) y = (x2 + 5)6

(e) y = ex (f) y = ln x (g) y = 2x (h) y = xx

11 [Points: 2+2+2+2+2] Which of the following statements are correct?(a) The rule which converts a temperature measured in degrees Fahrenheit into the same temperaturemeasured in degrees Celsius is an invertible function.(b) A concave function always has a maximum.(c) A differentiable function can only have an interior maximum at a stationary point for the function.(d) If f ′(a) = 0, then a is either a local maximum point or a local minimum point.(e) The conditions f ′(a) = 0 and f ′′(a) < 0 are necessary and sufficient for a to be a local maximumpoint for f .

12 [Points: 2+2+2+2] In each of the following cases, decide whether the given formula is correct or not:

(a)∫

x2 dx = 13x3 + C

(b)∫

[f (x) + g(x)] dx = ∫f (x) dx + ∫

g(x) dx

(c)∫

f (x)g(x) dx = ∫f (x) dx

∫g(x) dx

(d)∫ b

ax dx = b − a


86

Answers to TEST I

1. (a)73 · 72

74= 73+2

74= 75

74= 75−4 = 71 = 7 (b) (5.5 − 3.5)3 = 23 = 8

(c)

(−2

5

)(−2

5

)(−2

5

)= −8

125(= −0.064) (d)

219 − 217

219 + 217= 217(22 − 1)

217(22 + 1)= 3

5

2. (a) 4x4y2 = (2x2y)2 = 25 (b) 11 % of 3500 is 3500 · 11/100 = 3500 · 0.11 = 385(c)

√132 − 122 = √

(13 + 12)(13 − 12) = √25 = 5 (or

√132 − 122 = √

169 − 144, etc.)(“

√132 − 122 = √

132 − √122 = 13 − 12 = 1” is a SERIOUS mistake.)

(d)

√3 + √

2√3 − √

2= (

√3 + √

2)(√

3 + √2)

(√

3 − √2)(

√3 + √

2)= 3 + 2

√3√

2 + 2

3 − 2= 5 + 2

√6

3. (a) (x + 2y)2 = x2 + 4xy + 4y2 (b) (2x − 3y)2 = 4x2 − 12xy + 9y2

(c) (a + b)(a − b) = a2 − b2

4. (a) 5a − (3a + 2b) − 2(a − 3b) = 5a − 3a − 2b − 2a + 6b = 4b

(b) (x + 2)2 + (x − 2)2 − 2(x + 2)(x − 2) = [(x + 2) − (x − 2)]2 = 42 = 16(c) (1−x)2(1+x)2 = [(1−x)(1+x)]2 = (1−x2)2 = 1−2x2+x4 (d) (2−a)3 = (2−a)2(2−a) =(4 − 4a + a2)(2 − a) = 8 − 4a − 8a + 4a2 + 2a2 − a3 = −a3 + 6a2 − 12a + 8

5. 8(1.05)6 billion dollars.

6. (a) 5a2b + 15ab2 = 5ab(a + 3b) (b) 9 − z2 = (3 − z)(3 + z)

(c) p3q − 4p2q2 + 4pq3 = pq(p2 − 4pq + 4q2) = pq(p − 2q)2. (One point for the first equality.)

7. (a)1

2− 1

3= 3

2 · 3− 2

3 · 2= 3

6− 2

6= 1

6(b) 20 is the common denominator, so

6a

5− a

10+ 3a

20= 4 · 6a

20− 2a

20+ 3a

20= 24a − 2a + 3a

20= 25a

20= 5a

4

(c)12 − 1

314 − 1

6

=6

12 − 412

312 − 2

12

=2

121

12

= 2 (or 12 − 1

3 = 2( 14 − 1

6 ), so the ratio is 2).

8. (a) 251/2 = 5 (b) (x1/2y−1/4)4 = x(1/2)·4y(−1/4)·4 = x2y−1 (c) 3√

27a6 = 3√

27 3√

a6 = 3a2

(d) p1/5(p4/5 − p−1/5) = p1/5p4/5 − p1/5p−1/5 = p1/5+4/5 − p1/5−1/5 = p − 1

9. (a) 3x = −30, so x = −10(b) 2x + 3 = 3(x − 1), so x = 6. (Neither denominator is 0 when x = 6.)(c) If

√3 − x = 2, then 3 − x = 4, so x = −1. This is indeed a solution, as is easily checked.

10. (a) −3x + 2 < 5 or −3x < 3, so that x > −1. (Recall that an inequality is reversed if multipliedby a negative number.)(b) −3 < x ≤ 1. (Use a sign diagram. Note that the fraction is undefined if x = −3.)(c) x < −1 or 0 < x < 1. (x3 < x, or x3 − x < 0, and so x(x2 − 1) < 0, or x(x − 1)(x + 1) < 0.Then use a sign diagram.)

11. (a) 3x(1 − 3x) = 0, so x = 0 or x = 1/3 (b) x = −3, x = 5 (c) P = −2 or P = 1/2

12. (a) x = 3, y = 1 (b) p = 10, q = 2 (c) Put x = 1/p, y = 1/q. Then 3x + 3y = 3 and3x − y = 7. The solution to this system is x = 2 and y = −1. But then p = 1/2 and q = −1.


87

Answers to TEST II

1. (a) 25 + 25 = 2 · 25 = 26, so x = 6.

(b) 3−15 + 3−15 + 3−15 = 3 · 3−15 = 3−14, so y = −14.

(c)√

132 − 52 = √169 − 25 = √

144 = 12.

(d)226 − 223

226 + 223= 223(23 − 1)

223(23 + 1)= 23 − 1

23 + 1= 7

9, so z = 7.

2. (a) (i) −3/2 (ii) 2 (b) (i) y = −2x − 1 (ii) y = (−b/a)x + b

3.x −2 −1 0 1 2 3 4

y = −x2 + 2x + 4 −4 1 4 5 4 1 −4

The graph of the function is shown in the figure below.

y

x

y

x

y

−4

−3−2

−1

12

3

45

x−4 −3 −2 −1 1 2 3 4

y = −x2 + 2x + 4

4. (a) Minimum 4 for x = 2. (Follows from x2 −4x +8 = x2 −4x +22 +8−22 = (x −2)2 +4, or byusing calculus.) (b) Maximum 18 for x = 4. (Follows from −2x2+16x−14 = −2(x2−8x+7) =−2(x2 − 8x + 42 + 7 − 42) = −2[(x − 4)2 − 9] = −2(x − 4)2 + 18, or by using calculus.)

5. (a)2x3 − 3x + 10

x + 2= 2x2 − 4x + 5

(b)x4 + x

x2 − 1= x2 + 1 + x + 1

x2 − 1= x2 + 1 + 1

x − 1

6. f ′(x) = 4x2 − x4 = x2(4 − x2) = x2(2 − x)(2 + x) = 0 for x = 0 and for x = ±2.

7.y

x

y = f (x)

(a)

y

x

y = f (x)

(b)

y

x

y = f (x)

(c)

y

x

y = f (x)

(d)


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8. (a) The marginal cost is 50 when output is 1000 tons. (Or: The cost of extracting one ton more than1000 tons is approximately 50 dollars.) (b) P ′(t) ≤ 0, because more search never leads to a higherbest price, and generally leads to a lower one. However, P ′′(t) is likely to be positive, because thelonger you search the smaller gain in price you will obtain. (P(t) is decreasing and convex.)

9. (a) y = f (x) + g(x) ⇒ y ′ = f ′(x) + g′(x)

(b) y = f (x)g(x) ⇒ y ′ = f ′(x)g(x) + f (x)g′(x)

(c) y = f (x)/g(x) ⇒ y ′ = f ′(x)g(x) − f (x)g′(x)

(g(x))2

(d) y = f (g(x)) ⇒ y ′ = f ′(g(x))g′(x)

10. (a) y = x2 ⇒ y ′ = 2x (b) y = 15x5 ⇒ y ′ = x4 (c) y = x

x + 1⇒ y ′ = 1

(x + 1)2

(d) y = (x2 + 5)6 ⇒ y ′ = 6(x2 + 5)52x = 12x(x2 + 5)5 (e) y = ex ⇒ y ′ = ex

(f) y = ln x ⇒ y ′ = 1/x (g) y = 2x ⇒ y ′ = 2x ln 2 (h) y = xx ⇒ y ′ = xx(ln x + 1)

11. (a) Correct. (b) Wrong. (Consider y = −ex .) (c) Correct. (d) Wrong. (a could be an inflectionpoint.) (e) Wrong. (The conditions are sufficient, but not necessary. For example, f (x) = −x4

has a maximum at x = 0 and yet f ′(0) = f ′′(0) = 0.)

12. (a) Correct. (b) Correct. (c) Wrong, because ddx

(∫f (x) dx

∫g(x) dx

) = f (x)∫

g(x) dx +g(x)

∫f (x) dx �= f (x)g(x), except for very special functions f and g.

(d) Wrong, because∫ b

ax dx = ∣∣b

a12x2 = 1

2 (b2 − a2), which = b − a only when a + b = 2.


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