CHAPTER 6

PAGE 1Chapter Five: Elements and Applications of Calculus

Chapter 5elements and applications of calculus

Calculus is the Branch of mathematics that concerns it self with the rate of change of one quantity with respect to another quantity.6.1Differential Calculus- Broadens the concept of slope - rates of change

There are two core concepts which lie down the foundation for differential calculus. These are: Limits and continuity.

Limits: In calculus there is often a concern about the limiting value of a function as The independent variable approaches some specific value, a, this limiting value when it exists is called a limit.

Notation:Lim f (x) = L

X(aA function f has a limit L (where L is some real number) as x approaches a if the values of the dependent variable f (x) differ arbitrarily little from L for all values of X which lie very close to a. The limit as X approaches a is symbolized:

Lim f (x) = L

X(a N.B.

The concept of the limit of a function as x approaches a should not be confused with the concept of the values of a function at x = a

The limit as x approaches a may exist, and the function many or may not be defined at a.

The function may be defined at a, and the limit may or may not exist.

The limit as x approaches a may exist, and the function may be defined at a and their values may, or may not be the same.

Generally, x can approach a from either of two directions, through values that are less than a or through values that are greater than a.

The limit L must be a finite number.

The limit f (X) does not depend on the values of f (x) at X = a. Whether X(a

or not the function, f(x) is defined at x = a doesnt affect the limit or its existence or non existence at x = a. In other words, lim f (x) = L

x(a means that the values of f (x) approach, with out necessarily being equal to L, as X approaches, without necessary being equal to, a

E.g. Left-Hand and Right-Hand LimitsWhen the variable x approaches the value x=a but always remains less than a, we say x is approaching a from the left. If the values of f(x) gets closer and closer to a real number L as x approaches a from the left, we call L the LEFT-HAND LIMIT and employ the symbolism

Lim f(x) = L( X(a(Only those values of X to the left of a are used to compute the left hand limit.

E.g. Lim (x+3)

X(3(We might compute f (2.9) = 5.9, f(2.99) = 5.99, -------------- f (2.999---) = 5.999 --------. Clearly, the values of f (X) are getting closer and closer to the number 6, and we write.

L- = lim (x+3) = 6

X(3(Similarly, the RIGHT-HAND LIMT of a function f as x approaches a is the value L+, which f(X) converges on as x approaches the point X = a from the right, but always, remains greater than a. We symbolize this limit as

Lim f(x) = L+ x(a+Only those values of X to the right of a on the number line are used to obtain the right - hand limit. We might compute the limit

L+ = lim (x+3) = 6

x(3+

by evaluating f (3.1), = 6.1 f(3.01) = 6.01 f(3.001) = 6.001 ----, f (3.00001) = 6.000001.

A function f has a limit L as X approaches a only if the left - hand and the right-hand limits are equal. Their common value is the limit of f as x approaches a Here (L- = 6) = (L+ = 6) = L , and we write lim (x + 3) = 6

x(3

Exercises

Find the limits for the following functions; if it exists

1. Lim (X2 - 1)

x(2

X 1.9(1.99(1.999(1.9999 (0 2.0001(2.001(2.01( 2.1

F(x)2.61(2.9601(2.996001(2.99960001(3.0040001(3.004001( 3.0401 - 3.41

L-

3 , L+= 3

Lim (x2 - 1) = 3

x(2

2. Lim

x-0.1-0.01-0.0010.0010.010.1

X(0 f(x)-1-1-111

L+ is not equal to, so Limit does not exist.

3. Lim

x-0.1-0.01-.0000010.0010.010.1

X(0 f(x)(((0.0320.10.32

Lim f(x) = 0

x(0In such case priority should be given for the domain to determine either the right/ left hand limit is relevant. For lim

, since the values of x are restricted to non- x(0negative numbers, the left hand limit is not relevant.

4. F(x) Lim f(x) =?x(4

Lim f (x)x3.93.993.9993.99994.0014.014.1

X(4 f(x)0.10.010.0010.00010.0010.010.1

L- = 0

L+ = 0=L

Lim f(x) = 0 limit exist; but the function doesnt exist at 4 because the point x(4 is not in the domain of the fun.

Limit PropertiesAssuming that the two limits exist:1. Limit of a Constant The limit of a constant function is the constant value.

2. Limit of a Sum or a Difference

The limit of a sum or difference is the sum or difference of the individual limits, provided that these limits exist.3. Limit of a Constant Times a Function

The limit of a constant times a function is the constant times the limit of the function, provided that the limit exists.

4. Limit of a Function to a Power

The limit of a function raised to a power is the power of a limit provided that the limit does exist.5. Limit of a Product

The limit of a product is the product of the limits provided that these limits exist.6. Limit of a Quotient

The limit of a quotient is the quotient of the limits provided that the limits exist and the denominator is not zero. Continuity of a FunctionThe continuity of a function can be determined at a specific point or over an interval.Continuity at a point

A function is said to be continuous at a point when there exists no gap at that point. At the point of discontinuity we cant find marginal functions (MR, MC---)

A function f is continuous at x = a if and only if all these conditions apply to f at a:

1. f(a) is defined (that is, the domain of a f includes x = a)2. Lim f(x) exists

x(a

3. Lim f(x) = f (a) whether x approaches a from the left or right.

X (a

Geometrically, a function, f is seen to be continuous at a point x=a when there is nether a hole nor a gap in the graph of f at x = a

If any one of the three requirements specified above is not satisfied, the function is f is said to be discontinuous at x = a.

Four cases of discontinuity

1. A function f may be discontinuous because the value of f at x = a is not defined, although the limit of f as x approaches a exists.

E.g. f(x) =

is not defined at x=2 and the fun is discontinued at x=2

2. A function f may be discontinuous because the limit as x approaches a does not exist.

E.g. F(x) =

L+ is different from L-

The limit doesnt exist as x approaches a if the function has different right hand and left hand limits at that point.

The graph of the function has a vertical gap at x=5. Such a vertical gap always indicates discontinuity of the function at this point. This type of discontinuity is often called JUMP DISCONTINUITY.

3. A function may be discontinues because the limit of f as x approaches a is not the value of f at x=a.

E.g. f(x) =

is discontinuous at x = 3. This is because lim f(x) = 3 is not equal to f (3) = 1

x(3

4. Another type of discontinuity exists. A function is said to have an infinite discontinuity at x=a if f(x) becomes arbitrarily large positive or negative as x(a- or x(a+.

E.g. f(x) = 1/xlim 1/x = -

and lim 1/x = +

hence

x(0-

x(0+Lim 1/x doesnt exist, nor is f (0) defined.

Continuity of a function over an interval

A function is said to be continuous over an interval if its graph has no Break, jumps, or holes in that interval inclusive [ ] or exclusive ( ). Alternatively, a continuous function has a graph which can be drawn without lifting the pencil from the paper over that interval.

A function f is continuous over an open interval if it is continuous at each number in that interval.

A function f is continuous on a closed interval [a, b] provided the following conditions are satisfied:

1. f is continuous over the open interval (a, b)

2. f (x) (f (a) as x(a from with in (a, b)

3. f (x) (f(b) as x(b from with in (a, b)

Continuity properties

1. Constant functionIf f(x) = K where K is a constant, then f(x) is continuous for all x. In other words, a constant function is continuous for all values of x.

2. Power functions

Functions of the form f(x) = xn and g(x) =, where n is a positive integer, are continuous over all values of x of their respective domains.3. Sum, difference and product

If f(x) and g(x) are continuous at a point, then f(x) + g(x), f(x) - g(x), and f(x) .g(x) are continuous over that point.4. Quotient

If f(x) and g(x) are continuous at a point, then f(x)/g(x) is continuous at that point provided that g(x)(0 at that point.Any polynomial function is continuous at all values of x. any rational function is continuous at all values of x, where its denominator does not equal to zero.Exercises

1. Technic, Inc. manufactures electronic circuitry for computers. For a particular unit, there is a variable cost of Birr 4 per unit and a fixed cost of Birr 8,000 for the first 10,000 units produced. If the number of units manufactured exceeds 10,000, the fixed cost increases by Birr 1,000.a. Define the cost function. Answer:

b. Where is the cost function discontinuous? Answer: Since the graph has a Break at x= 10,000, then C(x) is discontinuous at x=10,000. 2. One million Birr is deposited in to a savings account for 1 year at 12% compounded quarterly. If interest is added at the end of each quartera. Find the account's balance for each quarter. Answer: Birr 1,030,000, 1,060,900, 1,092,720, and 1,125,508.81. b. Where is the graph discontinuous? Answer: The graph is discontinuous at the end of the first, second, third and fourth quarters.

Continuity and Differentiability

The relationship between continuity and differentiability is that: A function differentiable at x=a is continuous at x=a. that is, differentiability implies continuity. But the converse of the preceding statement is not true; continuity doesn't imply differentiability. In other words, it is possible for a continuous at x=a but not differentiable at x=a.Intuitive development of the concept of derivative

In studying the relationship between variables our primary concern would be to observe the manner in which the value of one variable (the dependent variable) is changed with given increases or decreases in the value of the other variable (the independent variable). This analysis of how the value of the dependent variable responds to changes in the value of the controllable independent variable leads us to a study of RATES OF CHANGE and eventually to the study of DERIVATIVES. For a linear function each one unit change in x has the same effect on the value of y, no matter what the value of x is, we call this slope and remains constant for a linear function, whose graph is a straight line. This property is not the case for non-linear functions, i.e., the slope changes from one value of x to another. Hence, when the relationship between x and y is depicted by a non-linear function, the effect that a one-unit change in one unit of x has up on the value of y is it self a function of x. Because the slope is not a constant, it is not logical to speak of the slope of non linear functions as we did for a linear function. We instead may speak of the AVERAGE RATE OF CHANGE Between TWO POINTS (ARC), or INSTANTANEOUS RATE OF CHANGE AT A POINT (IRC)

Average pate of change (ARC) between two points

Average Rate of Change (ARC) of a function f over an interval x to x+(x is given by the change in f(x) divided by the change in X; that is,

ARC =

E.g. At the same instant that a test driver begins his journey around a truck, a stopwatch is started. The function is given by Y = f(x) = 10x2 (x>0) expresses the total distance, y (in miles) traveled by the driver during the first x hours. Thus, during the first 3 hrs, the driver has traveled a total distance of Y= f (3) = 10(3)2 = 90 miles.

During the first 5 hours, the driver has traveled a total distance of Y = f(5) = 10(5)2 = 250 miles

We now pose the following question:

What is the driver's average speed during the time interval between the end of the third hour and the end of the fifth hour?

Since the driver has traveled 90 miles during the first 3 hours and 250 miles during the first 5 hours, he has traveled

(y = 250 - 90 = 160 miles during the time interval (x = 5 - 3 = 2. Dividing by the length of the time interval, we have

as the average speed or the average rate of change of distance with respect to time. Note that the average speed is the slope of the straight line, L, passing through points (3,90) and (5,250)of the graph of Y = f(x)=10x2. Such a straight line intersecting the graph in at least two points is called a secant line.In general the slope of the secant line passing through two points of the graph of a function is the average rate of change of that function over the respective interval. Note that the secant line passes through the points (x,f(x)) and (x+(x, f(x+(x)). Thus, the average rate of change of the function Y= f(x) over the interval from x to x+(x is given by the expression

ARC =

This ratio is called a difference quotient. It gives the slope of the secant line passing through (x, f(x)) and (x+(x, f(x+(x)).The general procedure for determining a formula for the difference quotient is: 1. Replace x with x+(x to obtain f(x+(x))

2. Subtract f(x) from the result of step 1 to obtain (y = f(x+(x)- f(x)

3. Divide the result of step 2 by (x (and simplify) to obtain a formula for difference quotient. This result gives a formula for the average rate of change for the given function f(x) as we move from x to x+(x. Example 1. Compute the difference quotient of the function defined by Y = f(x) = 10x2

Step 1. Replace x with x+ (x f (x+(x) = 10(x+(x )2

= 10[x2 +2x((x) + ((x) 2]

= 10x2 +20x((x) + 10((x) 2

Step 2. Subtract f(x) from the result of step 1

= f(x+(x) - f(x)

= 10x2 +20x((x) + 10((x) 2 - 10x2

= 20x((x) + 10((x) 2

Step 3. Divide the result of step 2 by (x

= 20x + 10(xUsing this general formulation, we can determine the ARC for f(x) = 10x2 between any two points.

Exercises

1. The total revenue R, gained from selling x units of a product is given by R(x) = -0.2x2 + 8000x (0 0 the first derivative f ' is an increasing function at X.2. If f '' (x) < 0 the first derivative f ' is a decreasing function at X.3. If f '' (x) = 0 the first derivative f ' is a constant function at X.

Thus, the first and second derivatives taken together tell us a great deal more about the function than does the first derivative alone.

a. If f ' > 0 and f '' > 0, the function f is increasing at an increasing rate. That is, the slope of f is positive and is getting steeper. b. If f ' < 0 and f '' > 0, the function f is decreasing at an increasing rate. That is, the slope of f is negative but is becoming less steep.c. If f ' < 0 and f '' < 0, the function f is decreasing at an increasing rate. That is, the slope of f is negative and is becoming steeper.d. If f ' > 0 and f '' < 0, the function f is increasing at a decreasing rate. That is, the slope of f is positive but is becoming less steep.e. If f ' > 0 and f '' = 0, the function f is increasing at a constant rate. That is, the slope of f is positive and linear.f. If f ' < 0 and f '' = 0, the function f is decreasing at a constant rate. That is, the slope of f is negative and linear. Extreme Values of a function

Geometrically, the maximum and/or minimum values of a function are the high and/or low points on the functions graph.

Extreme values of a function are divided in to

- Local/Relative extrema - Maximum & minimum

- Absolute extrema - Absolute maxima and Absolute minima

Local Extrema: - usually for open intervals

a. A function f attains a relative/local maximum at x=a if there exists an open interval I containing a such that f (a) > f(x) for all x in the interval I. That is to say, a function is considered to achieve a relative maximum at some value of the independent variable if the value of the function at that point is at least as large as it is for any other nearby points.

b. A function attains a relative minimum at x=a if there exists an open interval I containing a such that f(a) < f(x) for all x in the interval I. That is to say, a function is considered to achieve a relative minimum at some value of the independent variable if the value of the function at that point is at least as small as it is for any nearby point.

Because a relative maximum need not be the highest point for the entire graph but simply a point higher than other points in its immediate neighborhood, a function may have more than one relative maximum. Similarly, a relative minimum is minimal only in relation to adjacent points, so a function may have more than one relative minimum.

Absolute Extrema - usually for closed interval.

a. A function f attains an absolute maximum at x = a if and only if f (a) > f(x) for all x in the domain of f.

b. A function f attains an absolute minimum at x=a if and any if f (a) < f(x) for all in the domain.

A point [a, f (a)] is an absolute extremum only if it is the very highest or lowest, point on the entire curve.

When the domain of f is a closed interval

If the domain of a function is restricted to some closed interval, and if the function is continuous on that domain, the function necessarily has both an absolute maximum and an absolute minimum.

The absolute extrema of such functions may occur either at an end point or at an interior point of the closed interval, or possibly at both. Relative extrema, on the other hand, are conventionally considered to occur only at interior points of the domain.

When the domain of f is defined over an open interval (a, b), or if the domain is unrestricted, then f may or may not have an absolute maximum or an absolute minimum.

First Derivative Test for Relative Extrema

Relative extrema occurs only at critical points of a function.

A critical point of a continuous function f is a point (x*, f (x)) such that either

a. F '(x*) = 0 or,

b. F ' (x*) does not exist but f(x*) is defined

X* is the Critical Value of x.

In searching for relative extrema we need only consider critical values of x. Thus, the first step in identifying relative extrema is to determine the set of x which comprises candidates for extrema; that is we determine x* = {x/f'(x*) = 0 or f ' (x*) does not exist but f(x*) is defined. However, these are only candidates for extrema; they are not necessarily relative extrema. The next step, thus, is to determine the character of each member of the X* set. One method of determining the character of a critical point is known as the First Derivative Test.To the immediate left of a maximum point the function is always increasing (that is, f ' >0) and to the immediate right of a maximum point the function is always decreasing (that is, f '0). So, THE ALGEBRAIC SIGN OF THE first derivative ALWAYS CHANGES FROM -VE TO +VE AS THE FUNCTION TRANSVERSES A RELATIVE MIMUM POINT.

If the sign of the first derivative does not change as the function transverses a critical point, the point is neither a maximum nor a minimum.

Example:

Find the relative maxima and minima of the following functions, if they exist.1. f (x) = x2 - 4x +4

a. f '(x) = 2x - 4

b. Use f '(x) to determine the behavior of f

2x - 4 = 0

1. f ' (x) < 0 for x 0 for x > 0 - f ( when x > 0

x = 0 - candidate

+ve

+ve

The algebraic sign of f ' does not change as we move from left to right across the critical point; therefore, the critical point x* 0, is neither a maximum or nor a minimum.

3. f(x) = 1/x2 The first derivative can never be negative and the function is undefined at x = 0, so no critical value.

4. f(x)=x2/3f '(x)= . The first derivative can't be set equal to zero. Though the first derivative can never be equal to zero, f(x) is defined at x=0. So the critical value is x=0. Since f '(x) is negative for all values of x to the left of zero and positive for all values of x to the right of 0, x=0 is a relative minima.5. f(x)= x3+3x2-72x+9f '(x)= 3x2+6x-72, setting f '(x)= 0 and solving yields

3x2+6x-72=0

3(x2+ 2x-36) =0

3(x+6) (x-4) =0Thus, x=-6 and x=4 are critical values. Since there are no values of x at which f '(x) does not exist, these are the only critical values. Since f '(x)>0 for all nearby values of x to the left of x=-6 and f '(x) 0 for all nearby values of x to the right of x=4, then, by the first derivative test, a relative minimum exists at x=4. Computing the corresponding y values we have (-6,333) as relative maxima and (4,-167) as relative minima.

Second Derivative Test for Relative Extrema

We have seen that the second derivative f '' gives information about the rate of change of the first derivative f '. If f ''>0 at some point x = a, then f ' is an increasing function at that point, and the graph of f is Concave upward. Or if f " < o at some point x = a, then f ' is a decreasing function at the point, and the graph of f is concave down ward.

If an extremum occurs at a point X* where f is concave downward, that extremum is a maximum. If an extremum occurs it a point X* where f is concave upward, that extremum is a minimum. These facts lead to a second procedure for locating local extrema, a procedure referred to as the Second Order Derivative Test. The second derivative test for locating relative extrema proceeds as follows:

1. Find the first derivative f ' and the second derivative f "of the function f. 2. Locate the set of critical values of x.

3. Evaluate the second derivative f '' at each critical value x*

A. If f '' (x*) < 0, then f(x*) is a maximum value of f.B. If f '' (x*) > 0, then f(x*) is a minimum value of f.C. If f '' (x*) = 0, then the test is indeterminate - use 1st derivative test.

Example:Use the second derivative test to locate the relative extrema

a. f(x) = x2 - 4x - 45

1. f '(x) = 2x - 4, f ''(x) =2

2x = 4

2. X = 2 = critical value

3. 2> 0 - f '' (2) = 2 - the function is concave upward at this point, Hence, the point (2, f (2)) is a relative minimum point on function f.

b. f(x)= x3 +3x2 + 3x - 5

1. f '(x) = 3x2 + 6x + 3, f '' (x) = 6x + 6

2. 3x2 + 6x + 3 = 0

Use 1st derivative test

(3x+3) (x+1) = 0 x -3 -2 -1 123 3x=-3, x=-1

f '(x) 12 3 0122748 x = -1 Critical value

+ve

+veTherefore, No relative extremum.

c. f(x) = x2 - 6x + 7

1. f '(x) = 2x -6

f ''(x) =2

2. 2x - 6=0

x = 3 critical value.

3. f ''(x)= 2 >0, therefore, [3, f(3)] is a relative minimum.

Locating Absolute Maxima and Minima

In locating absolute extrema we must consider whether the function is defined in a closed interval [a, b] or an open interval (a, b). If the domain is not specified, it is generally considered to be the set of all real numbers.

When the domain of f is a closed interval- an interval that contains both of its endpointsA continuous function f defined on a closed interval [a, b] must attain both an absolute maximum and an absolute minimum at points in [a, b].

If f is continuous on [a, b] with critical values x*1, X*2, --------X*n at interior points of the interval, the point at which f attains its absolute maximum must be critical value of x which represents a relative maximum of at one of the end points of the interval a or b. Similarly, f must attain its absolute minimum either at one of the interior point critical values which represents a relative minimum or at a and b.

Thus, to find the absolute maximum and absolute minimum, we evaluate f(x*1), f(x*2) ------- f (X*n), f (a) and f (b). The largest of these is the absolute maximum of f on the interval [a, b]; the smallest of these is the absolute minimum of f on the interval [a, b]. The other critical values - the other X* - represent relative extrema. The end points a and b are considered for absolute extrema only, and not as candidates for relative extrema.

In short, to find the absolute maximum on a closed interval, we follow the following steps.

1. Find the x coordinates of all the first-order critical points of f in the interval a 0 and A < 0.Example

1. Find any relative maxima and minima of the function z = f(x, y) = x2 8x + y2 12y + 1500

Solution First, we find the critical points. Computing fx and fy gives

fx = 2x 8 and fy = 2y - 12

Setting fx and fy equal to zero and solving for x and y gives x = 4 and y = 6. Hence, (4, 6)is the only critical point. We now apply the second derivative test for functions of two variables. Computing fxx, fyy, and fxy gives

fxx(x, y)= 2, fyy(x, y)= 2 and fxy(x, y)= 0.

Evaluating each of these at the critical point (x, y) gives

A = fxx(x, y) = 2

B = fyy(x, y) = 2

C = fxy(4, 6) = 0AB C2= 2(2)-02 = 4Since AB C2> 0 and A > 0, then, according to the second derivative test, a relative minimum occurs at (4, 6). Thus, the relative minimum value of z is z (4, 6) = (4)2 8(4) + (6)2 12(6) + 1500 = 1448.To find relative extrema for functions of two variables1. Search for critical points

Find the first order partials fx and fy Set the first order partials equal to zero and solve for x and y

Determine all possible ordered pairs (x, y) that satisfy both equations fx(x, y) = 0 and fy(x, y) = 0. These are the critical points.2. Apply the second derivative test for functions of two variables Compute the second order partials fxx, fyy and fxy. Evaluate the second order partials at each critical point.

Apply the second derivative test to each critical point.

Example

1. The revenue, z, derived from selling x units of calculators and y units of adding machines is given by the function

Z = f(x, y) = -x2 + 8x - 2y2 + 6y + 2xy + 50

a) How many calculators and adding machines should be sold in order to maximize sales revenue?b) What is the maximum sales revenue?Solution Critical points

a) we first calculate fx and fy, as follows:fx = -2x + 8 + 2y

fy = -4y + 6 + 2xSetting fx and fy equal to 0, we have

0 = -2x + 8 + 2y

0 = -4y + 6 + 2x

Solving for this linear system for x and y, we obtain x = 11 and y = 7. Thus, the only critical point is (11, 7).Second derivative test

We calculate fxx = -2, fyy = -4, and fxy = 2.Since the partial critical point is (11, 7), then

A = f (11, 7) = -2

B = f (11, 7) = -4

C= f (11, 7) = 2.

AB C2 = -2(-4) - 22 = 4Since AB C2 > 0 and A < 0, then, according to the second derivative test, a relative maximum occurs at (11, 7). Thus, in order to maximize revenue, x = 11 calculators and y = 7 adding machines must be sold.

b) The maximum sales revenue is

Z = f (11, 7) = - (11)2 + 8(11) 2(7)2 + 6(7) + 2(11) (7) + 50 = Birr 115.2. Meditech, Inc. produces two products used in the dental industry. Each thousand units of product 1 sells for Birr 100, and each thousand units of product 2 sells for Birr 80. Meditechs analysts have determined that if x thousand units of product 1 and y thousand units of product 2 are produced, the total production cost is given by C(x, y) = 10x2 + 5y2 10xy 20x +5y +12a) Determine the equation for total sales revenue, R(x, y).b) Determine the equation for total profit, P(x, y).c) Determine the number of units of each product that should be produced in order to maximize total profit.Solution

a) The total sales revenue is given by R(x, y) = 100x + 80y

b) The total profit function is given by P(x, y) = R(x, y) C(x, y)

= 100x +80y (10x2 + 5y2 10xy 20x +5y +12)

= -10x2 - 5y2 + 10xy + 120x + 75y 12

c) computing Px and Py, we have

Px = -20x + 10y + 120Py = -10y + 10y + 75

Setting Px and Py equal to zero and solving for x and y yields the critical point, (19.5, 27). Computing Pxx, Pyy, and Pxy and applying the second derivative test for functions of two variables gives

Pxx(x, y) = -20

Pyy(x, y) = -10

Pxy(x, y) = 10Hence,

A = Pxx(19.5, 27) = -20

B = Pyy(19.5, 27) = -10

C = Pxy(x, y) = 10

AB C2 = (-20) (-10) 102 = 100Since AB C2 > 0 and A 0 and A < 0, then by the second derivative test, a relative maximum occurs at (75, 85). Thus, brand 1 should be priced at Birr 75 per case and brand 2 should be priced at Birr 85 per case in order to maximize total profit.

The maximum profit is P (75, 85) = -2(75)2 (85)2 + 2(75) (85) + 130(75) + 20(85) - 4100

= Birr 1625.Lagrange Multipliers

Some times we must optimize a function z= f(x, y), where x and y are constrained. As an example, consider a factory that burs two types of fuel: BF108 and BF109. The number of tons of pollutant exhausted by the factory in a year is given by

Z = f(x, y) = x2 +2y2 xy 279,990.

Where x is the amount (in thousands of gallons) of BF108 fuel used annually and y is the amount (in thousands of gallons) of BF109 fuel used annually. The factory uses a combined amount of 800 thousand gallons of fuel annually. We seek to determine how many thousands of gallons of each type of fuel should be burned annually in order to minimize the amount of pollutant exhausted.Since the factory uses a combined amount of 800 thousand gallons of fuel annually, then

x + y = 800

Mathematically, our problem is to

Minimize f(x, y) = x2 +2y2 xy 279,990.

Subject to the constraint g(x, y) = x + y =800

Such a problem may be solved by the method of Lagrange multipliers. In general, the method of Lagrange multipliers is used to solve the following type of problem. Maximize (or minimize) z = f(x, y)

Subject to g(x, y) = C, where c is constant. To use the method of Lagrange multipliers, we define a new function

F(x, y, ) = f(x, y) + (c g(x, y))Where is called the Lagrange multiplier and the function F is called the lagrangian function. Studying the lagrangian function, F, note that since g(x, y) = C, then C g(x, y) = 0, and the value of F will equal that of the original function, f. Thus, by finding the values of x, y, and that maximize (or minimize) F, we also find the values of x and y that maximize (or minimize) f(x, y) subject to the constraint equation g(x, y) = C.

Returning to our problem, we find

F(x, y, ) = f(x, y) + (c g(x, y))

= x2 + 2y2 xy 279,990 + (800 x y)We now find the critical values of F.

Calculating FX, FY, F, we have

FX = 2x y FY = 4y x

F = 800 x ySetting these equal to zero yields

2x y = 0

4y x = 0

800 x y = 0

Solving the first equation for , we have

= 2x y

= 4y - x

Equating these two expressions for yields

2x y = 4y - x

-5y = -3x

y =

Substituting for y in to the third equation 800 x y yields 800 x = 0. Solving for x, we obtain x = 500.

Substituting x = 500 in to y =, we have y =

We now substitute x = 500 and y = 300 in to either of the equations for . arbitrarily choosing the equation = 2x y we obtain = 2(500) 300 = 700. Thus, the function F has a critical point x= 500, y = 300, and = 700. Then we test the critical points for maxima and minima. The function F has only one critical point at (500, 300, 700), we conclude that (500, 300) produces a minimum for f. Hence,

Min f(x, y) = f (500, 300)

= (500)2 + 2(300)2 (500) (300) 279,990.

= 10 is the minimum number of tons of exhausted pollutant.Interpretation of Lambda, If x = x0, y = y0 and = 0 are the optimal solution values to the constrained optimization problem

Maximize (or minimize) z = f(x, y) subject to g(x, y) = C then it can be shown that where is evaluated at the optimal solution values.In other words, 0 is the rate of change of z with respect to c and thus measures the sensitivity of the optimal value of f to a change in c. returning to our original example, recall that 0 = 700. Since = 0 = 700, then each unit increase in c (i.e., each additional thousand gallons of fuel used) increases the optimal amount of pollutant exhausted by approximately 700 tons. 2'

96 sq inch

Ocean

[Type text]Page 1

_1103188646.unknown

_1104072509.unknown

_1105259130.unknown

_1105431960.unknown

_1167744695.unknown

_1167745032.unknown

_1167745279.unknown

_1167745393.unknown

_1167745157.unknown

_1167744789.unknown

_1105433416.unknown

_1167744547.unknown

_1167744618.unknown

_1126625303.unknown

_1126626278.unknown

_1126708802.unknown

_1126709078.unknown

_1127135170.unknown

_1127227437.unknown

_1127302521.unknown

_1127302575.unknown

_1127226923.unknown

_1127134888.unknown

_1127134991.unknown

_1126709196.unknown

_1126708870.unknown

_1126708891.unknown

_1126708905.unknown

_1126708659.unknown

_1126626416.unknown

_1126708567.unknown

_1126625325.unknown

_1126624649.unknown

_1126624705.unknown

_1126623506.unknown

_1126624000.unknown

_1105433474.unknown

_1105432174.unknown

_1105432444.unknown

_1105432155.unknown

_1105431055.unknown

_1105431640.unknown

_1105431707.unknown

_1105431423.unknown

_1105259154.unknown

_1105430666.unknown

_1105259144.unknown

_1104133301.unknown

_1104913744.unknown

_1104915085.unknown

_1104921012.unknown

_1104921099.unknown

_1104920594.unknown

_1104913784.unknown

_1104141568.unknown

_1104913235.unknown

_1104913701.unknown

_1104141656.unknown

_1104141977.unknown

_1104134137.unknown

_1104135288.unknown

_1104133394.unknown

_1104074294.unknown

_1104074704.unknown

_1104079190.unknown

_1104074607.unknown

_1104073813.unknown

_1104074242.unknown

_1104073704.unknown

_1103985071.unknown

_1104056969.unknown

_1104066433.unknown

_1104066607.unknown

_1104072151.unknown

_1104066547.unknown

_1104057797.unknown

_1104057857.unknown

_1104057114.unknown

_1104051556.unknown

_1104056825.unknown

_1104056903.unknown

_1104051810.unknown

_1104050306.unknown

_1104051475.unknown

_1103985763.unknown

_1103438971.unknown

_1103984547.unknown

_1103984693.unknown

_1103984792.unknown

_1103984599.unknown

_1103984330.unknown

_1103984414.unknown

_1103441073.unknown

_1103197457.unknown

_1103386688.unknown

_1103438857.unknown

_1103277164.unknown

_1103191261.unknown

_1103196243.unknown

_1103189522.unknown

_1103190089.unknown

_1096439411.unknown

_1102781592.unknown

_1103120981.unknown

_1103124062.unknown

_1103187130.unknown

_1103124188.unknown

_1103124017.unknown

_1103115664.unknown

_1103120482.unknown

_1103115780.unknown

_1103115243.unknown

_1103114881.unknown

_1097302100.unknown

_1097307395.unknown

_1097307500.unknown

_1097307632.unknown

_1097307792.unknown

_1097307461.unknown

_1097302213.unknown

_1096440434.unknown

_1097301645.unknown

_1096440398.unknown

_1096270984.unknown

_1096369354.unknown

_1096439237.unknown

_1096439270.unknown

_1096439085.unknown

_1096360342.unknown

_1096369221.unknown

_1096271030.unknown

_1096269762.unknown

_1096269972.unknown

_1096270545.unknown

_1096269898.unknown

_1096268138.unknown

_1096268302.unknown

_1096268010.unknown