Business Mathematics eBook PART 2

Business Mathematics eBook

PART 2

Decision Analysis

It's the job of a manager to make decisions when planning or running business. Making decision means

selecting one out of several possible actions. Of course, he/she will choose the "best" decision so he/she can

"expect" best outcome/payoff/result happens in the future. The result of this decision also depends on what

event occurred. For example you have three alternatives: starting a small shop, starting a big shop or doing

nothing (the last alternative is also one of the alternative decisions). The result of the shop is depend on what

economic condition will be (worse, stable or better). Intuitively, if you know in the future that the economic

condition will be better (stable) you will start a big (small) shop. You will do nothing if you know that the

economic condition will be worse.

Three types of decision making:

1. Decision making under certainty

When a decision maker know which event will be in the future after he/she makes a decision.

2. Decision making under uncertainty

When a decision maker doesn't know which event will be in the future after he/she makes a decision.

The decision maker don't know the probability of each event happen.

3. Decision making under risk

When a decision maker doesn't know which event will be in the future after he/she makes a decision.

The decision maker know/ can assume the probability of each event happen.

Newsboy Problem

A newsboy has to determine the appropriate number of newspapers to purchase. Each paper costs Rp. 2000

and is sold for Rp. 3500. At the end of the day, any unsold papers can be disposed of at Rp. 50 each. He has

three alternatives: Each day he buys 30 or 40 or 50 newspapers. He can make a decision on how many

newspapers should he buy every day but he cannot control on the demand of the newspapers. So, what

should he do?

Before making decision, the newsboy calculate of the possible outcomes for every decisions.

If he buys 30 newspapers and the demand is 30, his profit will be = 30(Rp 3500)-30(Rp 2000)= Rp

45000

If he buys 30 newspapers and the demand is 40, his profit will be = 30(Rp 3500)-30(Rp 2000)= Rp

45000

If he buys 40 newspapers and the demand is 30, his profit will be = 30(Rp 3500)+10(Rp 50)-40(Rp

2000)= Rp 25500

Please calculate the six other outcomes by yourself.

Decision demand=30 demand=40 demand=50

buys 30 45000 45000

buys 40 25500

buys 50

Decision demand=30 demand=40 demand=50

buys 30 Rp 45000 Rp 45000 Rp 45000

buys 40 Rp 25500 Rp 60000 Rp 60000

buys 50 Rp 6000 Rp 40500 Rp 75000

Decision under certainty:

If the newsboy know that the demand will be 30, he will buy 30 papers to maximize profit (he will not

buy 40 papers or 50 papers).





Decision Under Uncertainty:

His decision depends on the decision criteria.

Maximin (pessimist) criteria: select the decision with maximum payoff in worst events → buys 30

papers

Maximax (optimist) criteria: select the decison with maximum payoff in best events → buys 50 papers

Decision under risk:

Suppose, he assumes that the probability of demand=30 is 25%, demand=40 is 50% and demand=50 is 25%.

The expected value if he buys 30 papers is 25%(45000)+50%(45000)+25%(45000)=Rp 45000



Select decision wich gives maximum expected value → buys 40 papers

Differentiation

Given y=f(x) then the slope at point x=x0 is f '(x0) where f '(x) = derivative of f at x. The process of finding the derivative is called: Differentiation.

From above graph, the slope at point A = dy/dx where dx is very small. The slope at point A is positive (dx positive → dy positive). From point A to B, the the slopes is decreasing and at point B the slope is zero. The slope at point C is negative (dx positive → dy negative) and at point D is zero. Point B has the maximum y and D has the minimum y.

Common rules for differentiation

Rule y = f(x) y' = f '(x) Notes

1 y = c 0 c = constant number

2 y = c.xn c.n.xn-1 c, n = constant numbers

3 y = c.e(a.x+b) c.a.e(a.x+b) a, b, c = constant numbers. e=2.718281828=a base of natural logarithm (ln)

4 y = c.ln(a.x+b) c.a/(a.x+b) c, a, b=constant numbers

Examples a) y = 2x3 -3√x → y' = 6x2 -3(0.5)x -0.5 = 6x2 -1.5/√x

b) y = 5e2x -3/e4x → y' = 10e2x -3(-4)/e4x = 10e2x +12/e4x

c) y = 2e5x+2 → y' = 2(5)e5x+2 = 10e5x+2

d) y = 2ln(-3x+5) → y' = -6/(-3x+5) = 6/(3x-5)

First and second derivatives y' is the first derivative of y. y" is the second derivative of y. y" also the first derivative of y'

Finding turning points (max and min y) Using above graph, point B and D are turning points. Point B has maximum y and point D has minimum y. Here is the way to calculate the exact point coordinates (x,y) of B and D.

Given y = 0.07x3-1.6x2+7.64x+5

y' = 0.21x2-3.2x+7.64 y' = 0 Using ABC formula: xB=2.9641 and xD=12.2740 yB=0.07(2.9641)3-1.6(2.9641)2+7.64(2.9641)+5=15.4113 yD=0.07(12.2740)3-1.6(12.2740)2+7.64(12.2740)+5=-12.832 y" = 0.42x -3.2 At B, y" = 0.42(2.9641)-3.2 = -1.96 (negative) → Point B(2.9641,15.4113) is maximum point. At D, y" = 0.42(12.2740)-3.2 = 1.96 (positive) → Point D(12.2740,-12.832) is minimum point.

Try answer following questions

Given y=f(x)=3e(2x-5)-x√x+ln(10-2x) then

f(1)=

f(4)=

f '(1)=

f '(4)=

Given y=f(x)=(1/3)x3-(1/2)x2-6x+25 then

f(-3)=

f(4)=

y-intercept=

max point at x= and y=

min point at x= and y=

Given y=f(x)=(2/3)x3-2x2-6x+50 then

f(-2)=

f(4)=

y-intercept=

max point at x= and y=

min point at x= and y=

Application Differentiation

Marginal Revenue (MR)

Total Revenue = TR = Price (P) x Quantity sold

Given demand function: P=f(Q) then TR is also f(Q).

In economics textbooks, MR is defined as the

change in TR per unit change in Q → MR = ∆TR/∆Q

When TR is non linear, MR = ∆TR/∆Q ≈ d(TR)/dQ.

Example 1:

Given demand function: Q = 12-2P calculate the

value of TR & MR for Q=1,2,3,4,5,6,7

Solution:

Demand function: Q = 12-2P → P = 6 -0.5Q

Total Revenue: TR = P.Q = (6-0.5Q)Q = 6Q -0.5Q2

Marginal revenue: MR ≈ d(TR)/dQ = 6 -Q

Quantity sold Total Revenue Marginal Revenue

1 6(1)-0.5(1)2=5.5 6-1=5

2 6(2)-0.5(2)2=10.0 6-2=4

3 6(3)-0.5(3)2=13.5 6-3=3

4 6(4)-0.5(4)2=16.0 6-4=2

5 6(5)-0.5(5)2=17.5 6-5=1

6 6(6)-0.5(6)2=18.0 6-6=0

7 6(7)-0.5(7)2=17.5 6-7=-1

By looking at above table, we know that the

producer/seller will not sell more after selling 6

unit as the total revenue will decrease. Also

looking at MR at Q=6 is zero, we can conclude that

the by selling 6 unit will give TR maximum and

additional unit sold will decrease the TR.

Marginal Cost (MC)

Marginal cost is the cost of the next unit produced.

Given Total Cost = Fixed Cost + Variable Cost

Variable Cost = VC = f(Q)

MC = ∆TC/∆Q ≈ d(TC)/dQ = d(FC+VC)/dQ =

d(VC)/dQ = MVC

since the derivative of fixed cost (a constant) is

zero.

MVC = Marginal Variable Cost.

Example 2:

A craftsman has fixed costs of Rp 1,000,000 and a

cost of Rp 15,000 for each bracelet he produces.

a) Derive an equation for Total Cost

b) Derive an equation for Marginal Cost

c) Does Marginal Cost vary with output?

Solution:

a) Total Cost = TC = 1,000,000 + 15,000Q

b) Marginal Cost = MC = TC' = 15,000

c) Marginal Cost is constant, does not vary with

output.

Example 3:

Given the total cost function, TC=0.25Q3-

8Q2+100Q+120:

a) How much is the fixed cost?

b) Derive an equation for Marginal Cost

c) Does Marginal Cost vary with output?

d) Estimate the approximate change in TC as

output (Q) increases from 15 to 16 units.

Solution:

a) Fixed cost = FC = 120

b) Marginal Cost = MC = TC' = 0.75Q2-16Q+100

c) Marginal Cost does vary with output.

d) ∆TC ≈ MC(15) = 0.75(15)2-16(15)+100=28.75

Thus the approximate change in TC is 28.75

Note: The exact change in TC is 32.25. It can be

calculated from TC(16)-TC(15)

Let you answer following question:


8Q2+100Q+120

The minimum Marginal Cost happens when Q=

The minimum Marginal Cost is

Average Revenue (AR) and Average Cost (AC)

Average Revenue = AR = TR/Q

TR = P.Q → AR = P.Q/Q = P

Average Cost = AC = TC/Q

AC = (FC+VC)/Q = FC/Q + VC/Q = AFC + AVC

AFC=Average Fixed Cost and AVC=Average

Variable Cost

Example 4:

Given:

* Demand function: P=25 is applied for a perfectly

competitive firm.

* Demand function: P=50-0.625Q is applied for a

monopolist.

a) Draw a graph of functions P, AR, and MR for the

perfectly competitive firm.

b) Draw a graph of functions P, AR, and MR for the

monopolist.

c) What conclusions from a) and b)

Solution:

a) P=25

TR=25Q

AR=TR/Q=25Q/Q=25

MR=TR'=25

b) P=50-0.625Q

TR=P.Q=50Q-0.625Q2

AR=TR/Q=50-0.625Q

MR=TR'=50-1.25Q

c) * For a perfectly competitive firm, Average

Revenue and Marginal Revenue functions are

equal to the demand function.

* For a monopolist, Average Revenue function is

equal to the demand function.

* For a monopolist, the slope of Marginal

Revenue function is double that of the demand

function.

* For a monopolist, MR < AR

* For a monopolist, maximum output is 40 as

producing more than 40 will decrease the Total

Revenue (MR negative).

Let you answer following question:


8Q2+100Q+120

The minimum Average Variable Cost happens

when Q=

The minimum Average Variable Cost

is

Example 5:

Given the average cost function, AC=0.25Q2-

8Q+100+120/Q:

a) How much is the fixed cost?

b) Derive an equation for Marginal Cost?

a) AC=TC/Q → TC=AC.Q=0.25Q3-8Q2+100Q+120 →

Fix Cost=FC=120

b) MC=TC'=0.75Q2-16Q+100

Profit Maximization

Profit = TR - TC will be maximized when MR = MC

Example 6:

A demand function of corn is given by the equation

Pd = 45 -0.05Q.

The total cost of this commodity is given by the

equation TC = 0.0005Q3 -0.1Q2 +20Q +450.

a. Write down the Marginal Revenue function

b. Write down the Marginal Cost function

c. Find the maximum profit

Solution:

a. TR = P.Q = (45 -0.05Q)Q= 45Q - 0.05Q2 → MR =

TR' = 45 -0.1Q

b. MC = TC' = 0.0015Q2 -0.2Q +20

c. MC = MR → 45 -0.1Q = 0.0015Q2 -0.2Q +20 →

0.0015Q2 -0.1Q -25 = 0 → using ABC formula Q =

166.66667

→ Max Profit = 45*166.66667-

0.05*166.666672-(0.0005*166.666673 -

0.1*166.666672 +20*166.66667+450) = 2790.7407

Several Variable Functions and Partial Derivatives

Functions of several variables

You have learned that usually a dependent

variable is influenced by many independent

variables. For instance a demand function:

Q=f(P,Y,Ps,Pc,Ta,A,...). So far we simplify the model

into one independent variable (Q=f(P)). Now we

are going to learn about functions with two

variables.

Functions of two variables

General form: z=f(x,y)

Example: z = x2 +2xy +2y2 -16x-20y +40

Production function: Q = f(L,K)

Where Q=Quantity Output, L=Labor Input and

K=Capital Input

Example: Q = 80L0.5K0.2

Total Revenue function for two products

Example:

Given: Demand function for product A: QA =

12 - PA/3

Demand function for product B: QB =

14 - 0.25PB

Find the Total Revenue function!

PA = 12(3) -3QA = 36 -3QA

PB = 14/0.25 -QB/0.25 = 56 -4QB

TR = PAQA +PBQB = (36-3QA)QA +(56 -

4QB)QB = 36QA -3QA2 +56QB -4QB

2

Profit function for one product but sells it in two

separate markets with price discrimination

Example:

Given: Demand function in market 1: Q1 = 12 -

P1/3

Demand function in market 2: Q2 = 14 -

0.25P2

Cost function: TC = 180 +9Q where

Q=Q1+Q2

Find the Profit function!

P1 = 12(3) -3Q1 = 36 -3Q1

P2 = 14/0.25 -Q2/0.25 = 56 -4Q2

TR = P1Q1 +P2Q2 = (36-3Q1)Q1 +(56 -

4Q2)Q2 = 36Q1 -3Q12 +56Q2 -4Q2

2

TC = 180 +9Q1+9Q2

Profit = TR - TC = 27Q1 -3Q1

2 +47Q2 -

4Q22 -180

After we know the functions then the problem is

to find the value of the independent variables

which maximizes (or minimizes) the dependent

variable.

We need First-Order and Second-Order Partial

Derivatives!

First-Order and Second-Order Partial Derivatives

Given z = f(x,y) then

∂z/∂x = zx = The first partial derivative of z

with respect to (w.r.t.) x and treat other

variables as constant

∂z/∂y = zy = The first partial derivative of z

w.r.t. y and treat other variables as

constant

∂/∂x(∂z/∂x) = zxx = The straight second-

order partial derivatives w.r.t. x

∂/∂y(∂z/∂y) = zyy = The straight second-

order partial derivatives w.r.t. y

∂/∂x(∂z/∂y) = zxy = The mix second-order

partial derivatives (zxy = zyx)

Examples:

z = f(x,y) zx zy zxx zyy zxy

z = x2 +2xy +2y2 -16x-

20y +40

2x +2y -

16

2x +4y -

20 2 4 2

Find the optimum points of z=f(x,y)

1. Find zx, zy, zxx, zyy and zxy

2. Solve the two equations: zx = 0 and zy = 0

to find x and y of the turning points

3. Calculate z based on x and y of the

turning points

4. Calculate Δ = zxx(zyy) -(zxy)2

5. Determine the nature of the turning

point:

zxx and zyy Δ Conclusion

Positive and Positive Positive Minimum Point

Negative and Negative Positive Maximum Point

Same signs Negative Inflection Point

Different signs Negative Saddle Point

Examples:

For z = f(x,y) = x2 +2xy +2y2 -16x-20y +40

zx = 2x +2y -16

zy = 2x +4y -20

zxx = 2

zyy = 4

zxy =2

zx = 0 and zy = 0

2x +2y -16 = 0

2x +4y -20 = 0

0 -2y +4 = 0 → y = 2

2x +2(2) -16 = 0 → x = (16-4)/2 = 6

z = 62 +2(6)(2) +2(2)2 -16(6)-20(2) +40 = -28

Δ = 2(4) -22 = 4

As zxx > 0, zyy > 0 and Δ > 0 → A point (x,y,z) = (6,2,-

28) is a minimum point

Differential and small changes (incremental

changes)

The sign Δz ≅ means the change of z is

approximately equal

Example:

Given a Production function Output = f(Labor,

Kapital):

Q = 80L0.5K0.2

If L is increased by 1% and K is decreased by 2%,

use differentials to find approximate percentage in

Q.

Solution:

ΔQ ≅ 40L-0.5K0.2(1%L) + 16L0.5K-0.8(-2%K)

ΔQ ≅ 40%L0.5K0.2 -32%L0.5K0.2

ΔQ ≅ 8%L0.5K0.2

ΔQ ≅ 8%/80(80)L0.5K0.2

ΔQ ≅ 0.1% Q

Thus, if Labor is increases by 1% and Kapital is

decreased by 2% then Output will be increased by

0.1%.

Maximizing Profit for one product but sells it in

two separate markets with price discrimination

Example:

Given: Demand function in market 1: Q1 = 12

- P1/3

Demand function in market 2: Q2 = 14

- 0.25P2

Cost function: TC = 180 +9Q where

Q=Q1+Q2

a) Find

the

quantities in each market which

maximizing the profit.

Using

second partial derivatives, make

confirmation that the quantities will

give maximum profit.

b) Find

the

prices in each market which

maximizing the profit

c) Find maximum profit

the

Solution:

a) P1 = 12(3) -3Q1 = 36 -3Q1

P2 = 14/0.25 -Q2/0.25 = 56 -4Q2

TR = P1Q1 +P2Q2 = (36-3Q1)Q1 +(56 -

4Q2)Q2 = 36Q1 -3Q12 +56Q2 -4Q2

2

TC = 180 +9Q1+9Q2

Profit = Π = TR - TC = 27Q1 -

3Q12 +47Q2 -4Q2

2 -180

ΠQ1 = 27 -6Q1 ΠQ2 = 47 -8Q2

ΠQ1Q1 = -6 ΠQ2Q2 = -8 ΠQ1Q2 = 0

From ΠQ1 = 0 and ΠQ2 = 0 → Q1 = 27/6

= 4.5 and Q2 = 47/8 = 5.875

As ΠQ1Q1 < 0, ΠQ2Q2 < 0 and Δ=(-6)(-

8)+02 = 48 > 0 then maximum point

confirmed

b) P1 = 36 -3(4.5) = 22.5

P2 = 56 -4(5.875) = 32.5

c) Max Profit = 27(4.5) -3(4.5)2 +47(5.875) -

4(5.875)2 -180 = 18.8125

Integration and Application

Integration as the reverse of differentiation

Let y = f(x) = x2/2 +c where c=constant

number)

We have learned that y ' = f '(x) = dy/dx = x

Now we learn that ∫ (x)dx = x2/2 +c

Generic Integration Formula

Some examples

∫ (x3)dx = 1/4 x4 +c

∫ (e2x+3)dx = 1/2 e2x+3 +c

∫ (2x3 -1/x)dx = 2/4x4 -ln|x| +c

Notes: |x| means absolute number of x

Integration by Algebraic Substitution

Some times we cannot just use the generic

integration formula.

For example: y = f(x) = (5x-2)10

Find ∫ f(x) dx

The solution:

Let u = 5x-2 → du/dx = 5 → dx = du/5

Integration and Area

Let F(x) = ∫ f(x) dx

Then Area = F(b) - F(a)

Example 1:

Given y=-1+0.8x and xD=5

Find the area of CDB using integration

Answer:

xC=1/0.8=1.25

F(x)=∫ (-1+0.8x)dx=-x+0.4x2

F(5)=-5+0.4(5)2=5

F(1.25)=-1.25+0.4(1.25)2=-0.625

Area CDB=F(5)-F(1.25)=5-(-0.625)=5.625

Attention: Area below x-axis has negative sign.

Example: Area AOC=F(1.25)-F(0)=-0.625-0=-0.625

Example 2:

Given y=(2/3)x3 -2x2 -6x + 50

Calculate the shaded area (between x = -2

and x = 4)

Answer:

F(x)=

F(4)=(1/6)44-(2/3)43-(3)42+(50)4=152

F(-2)=(1/6)(-2)4-(2/3)(-2)3-(3)(-2)2+(50)(-2)=-

104

The Shaded Area=F(4)-F(-2)=152-(-104)=256

Try following problem:

Given y=f(x)=(1/3)x3-(1/2)x2-6x+25

and the plot:

Calculate the shaded area (between x=-3 to x=4)

Area=

Given f(x)=50-x2 and g(x)=10+6x, calculate the yellow area:

Area=

Consumer and Producer Surplus

In part one we have learned this topic.

The difference is the demand and supply functions are

non linear.

In this case, use integration to find the area between

curve and horizontal axis.

Consumer Surplus=The integration -Pe.Qe

Producer Surplus=Pe.Qe -The integration.

Example:

The demand function for a good is given as Pd = 34.5866 +40 / e0.2Q.

The supply function for this good is given as Ps = 10 + 2Q + 0.1Q2.

Given equilibrium quantity=10, calculate the consumer and producer surplus at equilibrium price.

Solution:

Qe=10 → Pe=Pe=10 + 2(10) + 0.1 (10)2=40

CS=34.5866(10)-200e-2 -(-200)-40(10)= 118.7988

The consumer surplus at Pe is 118.7988

∫ (10+2Q+0.1Q2)dQ = 10Q+Q2+0.033333Q3

PS=40(10)-(10(10)+102+0.03333(10)3 -0)=166.6667

The producer surplus at Pe is 166.6667

Solve differential equations

An equation y has dy/dx = 6x -2 and when x=3 then

y=0

Calculate y when x=1

Solution:

dy/dx=6x-2

∫ (dy/dx)dx = ∫ (6x-2)dx

y=3x2-2x+c

Given x=3 and y=0 → 0=3(3)2-2(3)+c → c=6-27=-21

y=3x2-2x-21

When x=1 then y=3(1)2-2(1)-21= -20

From Marginal Cost to Total Cost

Given MC=dTC/dQ then TC= ∫ (MC)dQ

The constant of integration c is the fixed costs.

Example:

Given the equation for marginal costs: MC = Q2 –

6Q + 11 and

TC = 500 when Q = 0

a) Find the equation for TC.

b) Find TC when Q=15.

Solution:

a) TC= ∫ (Q2 – 6Q + 11)dQ

= Q3/3 -3Q2 +11Q +c

500 = c

Total Cost = TC = Q3/3 -3Q2 +11Q +500

b) When Q=15 then TC= (15)3/3 -3(15)2 +11(15)

+500 = 1115

From Marginal Revenue to Total Revenue

Given MR=dTR/dQ then TR= ∫ (MR)dQ

The constant of integration c is usually equal zero.

Example:

Given the equation for marginal revenue: MR = 80

– 4Q

and TR = 0 when Q = 0,

find the equation for TR

Solution:

TR= ∫(80-4Q)dQ= 80Q-2Q2 +c

TR = 0 when Q = 0 → c=0

The total revenue function is: TR = 80Q-2Q2

Problem 1

The marginal price of a supply level of x bottles of

baby shampoo per week is given by

Find the price-supply equation if the distributor of

the shampoo is willing to supply 75 bottles a week

at a price of $1.60 per bottle?

Answer:

p=∫p.dx=∫300(3x+25)-2dx

Let u=3x+25 → du/dx=3 → dx=du/3

p=300∫u-2du/3 = 100∫u-2du = -100u-1+c= -

100/(3x+25)+c

1.6 = -100/(3(75)+25)+c → c = 1.6 +100/250 =2

The Price-supply equation is: P = 2 -100/(3x+25)

Problem 2

Paper and paperboard production in the United

States has steadily increased.

In 1990 the production was 80.3 (in million tons),

and since 1970 production has been growing at a

rate given by f ’(x) = 0.048x + 0.95

where x is years after 1970 and f(x) in million tons.

Find f(x), and the production levels in 1970 and

2000.

Answer:

f(x) = ∫f'(x)dx = ∫(0.048x + 0.95)dx = 0.024x2 +0.95x

+c

it's known when x = 1990 -1970 = 20 then f(x) =

80.3 → 80.3 = 0.024(20)2 +0.95(20) +c → c = 51.7

Thus f(x) = 0.024x2 +0.95x + 51.7

In 1970, production level = f(0) = 51.7 millions

In 2000, production level = f(30) =

0.024(30)2 +0.95(30)+ 51.7 = 101.8 millions

Problem 3:

A rate of consumption function of a product (in

thousands of litres per year) :

dQ/dt = 1560e0.012t

where t in years and Q=0 at t=0.

a) Calculate the total amount consumed during the

first 10 years.

b) Calculate the total amount consumed in the

years, 10 ≤ t ≤ 20

Answer:

Q = ∫(1560e0.012t)dt = (1560/0.012)e0.012t +c =

130000e0.012t +c

Q=0 at t=0 → 0=130000 +c → c= -130000

Q = 130000e0.012t -130000 = 130000(e0.012t -1)

a) At t=10 → Q = 130000(e0.12 -1) = 16574.59

The total amount consumed during the first 10

years is 16574.59 thousands of litres

b) At t=20→ Q = 130000(e0.24 -1) = 35,262.39

The total amount consumed in the years, 10 ≤ t

≤ 20 is 35,262.39 -16574.59 = 18687.8 thousands

of litres

Documents

Business Mathematics eBook PART 2