Dr. Karim Kobeissi

Chapter 3: Optimization Techniques

O p ti m i z a ti o n

Managerial economics is concerned with the

ways in which managers should make

decisions in order to optimize the

performance of the organizations they

manage.

Optimization Problems

An optimization problem involves the specification of three

things:

1) Objective function to be maximized (e.g. profit function) or

minimized (e.g., cost function).

2) Activities or choice variables (e.g., labor, quantity

produced, price, advertising…) that determine the value of

the objective function.

3) Any constraints that may restrict the values of the choice

variables (e.g., maximum total cost).

Optimum Level of Production

• Profit ()

Difference between total revenue (TR) and total cost (TC):

Profit = TR – TC

• Optimal level of production (Q*) is the level that maximizes

the profit.

TR

TC

Optimal Level of Production

1,000

Production Level

2,000

4,000

3,000

Q

0 1,000600200

Tota

l re

ven

ue a

nd

tota

l co

st

(dolla

rs)

Panel A – Total revenue and total cost curves

Q

0 1,000600200

Production Level

Pro

fit

(dolla

rs)

Panel B –Profit curve

•G

700

•F

••D’

D

•

•C’

C

•

•

B

B’

2,310

1,085

* = $1,225

•f’’

350 = Q *

350 = Q *

•M

1,225 •

c’’1,000

•d’’600

Marginal Benefit & Marginal Cost• Marginal Revenue(MR)– The change in total revenue (TR) associated with a one unit

change in the level of the independent variable (Q):

MR = DTR/DQ = Slope of Total Revenue Curve

• Marginal cost (MC)– The change in total cost (TR) associated with a one unit

change in the level of the independent variable (Q):

MC = DTC/DQ = Slope of Total Cost Curve

MC (= slope of TC)

MR (= slope of TR)

TR

TC

Relating Marginals to Totals

•F

••

D’

D

•

•C’

C

Production Level

800

1,000

Production Level

2,000

4,000

3,000

Q

0 1,000600200

Tota

l re

ven

ue a

nd

tota

l co

st

(dolla

rs)

Panel A – Measuring slopes along TR and TC

Q

0 1,000600200

Marg

inal re

ven

ue a

nd

m

arg

inal co

st (

dolla

rs)

Panel B – Marginals give slopes of totals

800

2

4

6

8

350 = Q *

100

520

100

520

350 = Q *

•

•

B

B’

b•

•G

•g

100

320

100

820

•

•

d’ (600, $8.20)

d (600, $3.20)

100

640

100

340

•

•c’ (200, $3.40)

c (200, $6.40)

5.20

Using Marginal Analysis to Find the Optimal Production Level

• If marginal benefit > marginal cost

– Production should be increased to reach highest net benefit

• If marginal cost > marginal benefit

– Production should be decreased to reach highest net benefit

• Optimal level of production

– When no further increase in profit is possible

– Occurs when MR = MC

Profit curve

Using Marginal Analysis to Find Q*

Q

0 1,000

600200

Production Level

Pro

fit

(dolla

rs)

800

•c’’

•d’’

100

300 100

500

350 = Q *

MR = MC

MR > MC MR < MC

•M

Differential Calculus in Management

• A function with one decision variable, X, can be written as:

Y = f(X)

• The marginal value of Y, associated with a small increase of X, is

My = DY/DX

• For a very small change in X, the derivative is written:

dY/dX = limit DY/DXDX B

Quick Differentiation Review

• Constant Y = c dY/dX = 0 Y = 5Functions dY/dX = 0

• A Line Y = c • X dY/dX = c Y = 5XdY/dX = 5

• Power Y = cXb dY/dX = b•c•X b-1 Y = 5X2

Functions dY/dX = 10X

Name Function Derivative Example

• Sum of Y = G(X) + H(X) dY/dX = dG/dX + dH/dX Two Functions

example Y = 5X + 5X2 dY/dX = 5 + 10X

• Product of Y = G(X) • H(X)Two Functions dY/dX = (dG/dX)H + (dH/dX)G

example Y = (5X)(5X2 ) dY/dX = 5(5X2 ) + (10X)(5X) = 75X2

Quick Differentiation Review

Marginal , S lope, Der ivati ve

• The marginal at point C is DY / DX

• The slope of the curve at point C is

equal to the slope of the tangent to the

curve at point C = the rise (DY) over

the run (DX) = DY / DX.

• Or the marginal at a point is equal to

the derivative at this point The

derivative at point C is also equal to its

slope:

Marginal = Slope = DerivativeX

C

DY

DY

DX

y

dy/dx 10

10 20

20

0

Max of Y

Slope = 0

value of x

Value of dy/dx whichIs the slope of y curve

Value of dy/dx when y is max

x

x

the function

Y = -50 + 100X - 5X2

i.e.,

dYdX

= 100 - 10X

if dYdX

= 0

X = 10

i.e., Y is maximized when

the slope equals zero.

Note that this is not sufficient for optimization problems.

In fact, the maximum / minimum points of a function

(e.g., profit function) occur when the slope of the

curve representing the function is equal to zero

Maximum / Minimum profit occur when the derivative

of the curve representing the profit function is equal

to zero.

Min value of y

Max value of y

d2y/dx2 > 0 d2y/dx2 < 0

value of dy/dx

y

dy/dx

x

x

Since dYdX

= 0 at two points, we need another

condition to distinguish between the maximum and

minimum points.

Look at the dYdX

curve

* at point 5 the curve is upward, i.e., its slope ( the

second derivative (the derivative of the derivative)) is

positive. Hence

d YdX

22

= > 0 ( minimum point )

* at point 10 the curve is downward, i.e., its slope is

negative. Hence

d YdX

22

= < 0 ( maximum point )

• Graphs of an original third-order function and its first and second derivatives. (what if the second order = 0)

Optimization Rules Maximization conditions:

1 - dYdX

= 0

2 - d YdX

22

= < 0

Minimization conditions:

1 - dYdX

= 0

2 - d YdX

22

= > 0

Applications of Calculus in Managerial Economics

Profit Maximization Problem

The profit function is ( = 50Q – Q2). The maximization of the function

occurs if :

1) First derivative [d/dQ = 50 - 2Q] at that point is equal to zero.

2) Second derivative [(d /dQ)’= -2] is <= 0.

Hence, Q = 25 will maximize profits.

More Applications of Calculus

Minimization problem: Cost Minimization

Supposes that there is a least cost point to produce. An average cost curve

might have a U-shape. At the least cost point, the slope of the cost

function is zero.

1) The first order condition for a minimum is that the derivative at that

point is zero.

2) The second order condition is that the second derivative is >= 0.

TC = 5Q2 – 60Q, then dC/dQ = 10Q – 60 and (dC /dQ )’ = 10.

Hence, Q = 6 will minimize cost

Where:

10Q - 60 = 0.

More Examples - Competitive Firm

• Maximize Profits () – Where = TR - TC = (P • Q)- (C • Q)– Use our first order condition: – d /dQ = 0 P-C = 0 PRICE = MC

Maximize = 100Q - Q2

First order = 100 -2Q = 0 implies Q = 50 and; = 2,500

Second Order Condition: one variable

• If the second derivative is negative, then it’s a maximum• If the second derivative is positive, then it’s a minimum

Max = 100Q - Q2

First derivative

100 -2Q = 0 second derivative is: -2 implies Q =50 is a MAX

Max= 50 + 5X2

First derivative

10X = 0second derivative is: 10 implies Q = 10 is a MIN

Problem 1 Problem 2

Partial Differentiation

• Economic relationships usually involve several independent variables.

• A partial derivative is like a controlled experiment- it holds the “other” variables constant

• Suppose price is increased, holding the disposable income of the economy constant as in

Q = f (P, I )

• then Q/P holds income constant.

Problem:

• Sales are a function of advertising in newspapers and magazines ( X, Y)

Max S = 200X + 100Y -10X2 -20Y2 +20XY

• Differentiate with respect to X and Y and set equal to zero.

S/X = 200 - 20X + 20Y= 0S/Y = 100 - 40Y + 20X = 0

• solve for X & Y and Sales

Solution: 2 equations & 2 unknowns

• 200 - 20X + 20Y= 0• 100 - 40Y + 20X = 0

• Adding them, the -20X and +20X cancel, so we get 300 - 20Y = 0, or Y =15

• Plug into one of them: • 200 - 20X + 300 = 0, hence X = 25• To find Sales, plug into equation:

• S = 200X + 100Y -10X2 -20Y2 +20XY = 3,250

PARTIAL DIFFERENTIATION AND MAXIMIZATION OF

MULTIVARIATE FUNCTIONS.

= f (Q1 , Q2 )

To know the marginal effect of Q1 on we hold Q2 constant, and

vice versa.

In order to do that we use partial derivative of with respect to

Q1 denoted by Q1( treating Q2 as constant )

e.g.;

= -20 + 100Q1 + 80Q2 - 10Q12 - 10Q2

2 - 5Q1Q2;

to find the partial derivative of with respect to Q1 we treat Q2

as constant; hence

Q1

= 100 - 20Q1 - 5Q2; (1)

therefore

Q2

= 80 - 20Q2 - 5Q1; (2)

setting both partial derivatives equal to zero and solve

simultaneously

100 - 20Q1 - 5Q2 =0

80 - 20Q2 - 5Q1 =0 multiply by -4 and add

________________

- 220 + 75Q2 = 0

hence

Q2 = 2.933

substitute for Q2 at any of the eq. 1

100 - 20Q1 - 14.665; hence

Q1 = 4.267.

i.e.,

profit is maximized when the firm produces 4.267 of Q1 and 2.933 of Q2.

CONSTRAINED OPTIMIZATION

We assume that the firm can freely produce 4.267 of Q1 and 2.933

of Q2. Quite often this may not be the case.

e.g.

Minimize TC = 4Q12 + 5Q2

2 - Q1Q2;

subject to:

Q1 + Q2 = 30 The constraint function

Solution:

The lagrangian multiplier:

Steps:

1 - set the constraint function to zero

2 - form the lagrangian function by adding the constraint function

after multiplication with an unknown factor to the original

function.

3 - take the partial derivatives and set them equal to zero

4 - solve the resulting equations simultaneously

step 1:

30 - Q1 - Q2 = 0

step 2:

L = 4Q12 + 5Q2

2 - Q1Q2 + ( 30 - Q1 - Q2)

step 3:

LQ1

= 8Q1 - Q2 -

2QL

= -Q1 + 10Q2 -

L

= -Q1 - Q2 + 30

8Q1 - Q2 - = 0 (1)

-Q1 + 10Q2 - =0 (2)

-Q1 - Q2 + 30 =0 (3)

step 4

multiply eq(2) by -1 and subtract from eq(1)

9Q1 - 11Q2 = 0 (4)

multiply (3) by 9 and add to eq(4)

-9Q1 - 9Q2 + 270 = 0

9Q1 - 11Q2 = 0

____________________

-20Q2 +270 = 0

Q2 = 270/20 = 13.5

substituting in eq (3) Q1 = 16.5

the values of Q1 and Q2 that minimizes TC are 16.5 and 13.5

respectively.

substituting Q1 and Q2 in eq(1) or eq(2) we find that

= 118.5

the interpretation of

measures the change in TC if the constraint is to be relaxed by one

unit.

i.e., TC will increase ( has a positive sign ) by 118.5 if the constraint

becomes 29 or 31.