Finance 30210: Managerial Economics

Optimization

Optimization deals with functions. A function is simply a mapping from one space to another. (that is, a set of instructions describing how to get from one location to another)

BAf :f

A

B

is a function

Is the range

Is the domain

A Bf

For example

For 53 xxf

Ax

50 x

5,0A ]20,5[B

3

14

Domain

205 y

Range

143 f

Function

y53 xy

For 50 x

0 5Domain

5

20

Ran

ge

X =3

Y =14

x

y 53max x50 x

0 5Domain

5

20R

ang

e

Optimization involves finding the maximum value for y over an allowable domain.

Here, the optimum occurs at x = 5 (y = 20)

x

y

x5

1max

100 x

What is the solution to this optimization problem?

105

There is no optimum because f(x) is discontinuous at x = 5

x

y x2max

60 x

0 6

12

What is the solution to this optimization problem?

There is no optimum because the domain is open (that is, the maximum occurs at x = 6, but x = 6 is NOT in the domain!)

x

y xmax

0x

0

12

What is the solution to this optimization problem?

There is no optimum because the domain is unbounded (x is allowed to become arbitrarily large)

x

The Weierstrass theorem provides sufficient conditions for an optimum to exist, the conditions are as follows:

x

y

x

x

The domain for is closed and bounded

is continuous

over the domain of

xf )(max xf

x

xfxxfxf

x

)()(lim'

0

xfFormally, the derivative of is defined as follows:

All you need to remember is the derivative represents a slope (a rate of change)

Actually, to be more accurate, the derivative represents a trajectory

Finding maxima/minima involves taking derivatives….

y

xxx x

Graphically…

x

xfxxfslope

)()(

xf

xxf

x

xfxxfxf

x

)()(lim'

0Now, let the change in x get arbitrarily small

y

xx

xf

x

xf

*x

0' xf

x

xf

*x

0' xf

0' xf 0' xf 0' xf 0' xf

First Order Necessary Conditions

If *x

)(max xfx

0*)(' xf

)(min xfx

is a solution to the optimization problem

then

or

Useful derivatives

Exponents1)(')( nn nAxxfAxxf

x

AxfxAxf )(')ln()(

LogarithmsLinear Functions

AxfAxxf )(')(

Example:

4'

4)(

xf

xxf Example:

x

xf

xxf

12'

ln12)(

Example:

4

5

15'

3)(

xxf

xxf

An Example

Suppose that your company owns a corporate jet. Your annual expenses are as follows:

You pay your flight crew (pilot, co-pilot, and navigator a combined annual salary of $500,000.

Annual insurance costs on the jet are $250,000

Fuel/Supplies cost $1,500 per flight hour

Per hour maintenance costs on the jet are proportional to the number of hours flown per year.

Maintenance costs (per flight hour) = 1.5(Annual Flight Hours)

If you would like to minimize the hourly cost of your jet, how many hours should you use it per year?

Let x = Number of Flight Hours

x

xx5.1$1500$

000,750$min

0

xx

CostHourly 5.1$1500$000,750$

hrsxx

7075.1

000,75005.1$

000,750$2

First Order Necessary Condition

An ExampleH

ou

rly

Co

st (

$)

Annual Flight Hours

0*)('' xf 0*)('' xf

Slope is increasingSlope is decreasing

x

xf

*x

0' xf

x

xf

*x

0' xf

0' xf 0' xf 0' xf 0' xf

How can we be sure we are at a maximum/minimum?

For a maximization… For a minimization…

Let x = Number of Flight Hours

x

xx5.1$1500$

000,750$min

hrsxx

7075.1

000,75005.1$

000,750$2

First Order Necessary Conditions

Second Order Necessary Conditions

0000,500,1$

3

xFor X>0

Suppose you know that demand for your product depends on the price that you set and the level of advertising expenditures.

22 5.8.4010000,5, pApAApApQ

Choose the level of advertising AND price to maximize sales

When you have functions of multiple variables, a partial derivative is the derivative with respect to one variable, holding everything else constant

First Order Necessary Conditions

010),( pAApQp

06.140),( ApApQA

22

0,05.8.4010000,5max pApAAp

Ap

10

010

pA

pA

010),( pAApQp

06.140),( ApApQA

50

40

06.24

0106.140

06.140

A

p

p

pp

Ap

With our two first order conditions, we have two variables and two unknowns

pAApQp 10),(

ApApQA 6.140),(

its generally sufficient to see if all the second derivatives are negative…

01),( ApQpp

06.1),( ApQAA

0*)('' xf

How can we be sure we are at a maximum?

Practice Questions

1) Suppose that profits are a function of quantity produced and can be

written as 222010 QQ

Find the quantity that maximizes profits

2) Suppose that costs are a function of two inputs and can be written as

2122

2121 32434 XXXXXXC

Find the quantities of the two inputs to minimize costs

Constrained optimizations attempt to maximize/minimize a function subject to a series of restrictions on the allowable domain

0

,max,

x,y gtosubject

yxfyx

To solve these types of problems, we set up the lagrangian

yxgyxfyx ,,),(

Function to be maximized Constraint(s)

Multiplier

To solve these types of problems, we set up the lagrangian

xgxfx )(

x

0

)(x

xf

*x

)(x xf

We know that at the maximum…

0)( xx

0xg

yxgyxfyx ,,),(

Once you have set up the lagrangian, take the derivatives and set them equal to zero

0,,),(

0,,),(

yxgyxfyx

yxgyxfyx

yyy

xxx

First Order Necessary Conditions

Now, we have the “Multiplier” conditions…

0, 0, 0 yxgyxg

Example: Suppose you sell two products ( X and Y ). Your profits as a function of sales of X and Y are as follows:

)(1.2010),( 22 yxyxyx

Your production capacity is equal to 100 total units. Choose X and Y to maximize profits subject to your capacity constraints.

100 yx

The key is to get the problem in the right format

0100

)(1.2010max 22

0,0

yxtosubject

yxyxyx

The first step is to create a Lagrangian

)100()(1.2010),( 22 yxyxyxyx

Objective FunctionConstraint

Multiplier

yxg ,

yxf ,

)100()(1.2010),( 22 yxyxyxyx

Now, take the derivative with respect to x and y

02.20),(

02.10),(

yyx

xyx

y

x

First Order Necessary Conditions

“Multiplier” conditions

0)100( 0100 0 yxyx

02.20

02.10

y

x

0 0100 yx

First, lets consider the possibility that lambda equals zero

0100 yx

50

02.10

x

x

100

02.20

y

y

05010050100 Nope! This can’t work!

02.20

02.10

y

x

The other possibility is that lambda is positive

0

0100 yx 0100 yx

x2.10 y2.20

50

50

2.202.10

xy

xy

yx

75

25

0250

050100

0100

y

x

x

xx

yx

yx 2.202.10

25

75

x

y05

Lambda indicates the marginal value of relaxing the constraint. In this case, suppose that our capacity increased to 101 units of total production.

Assuming we respond optimally, our profits increase by $5

Example: Postal regulations require that a package whose length plus girth exceeds 108 inches must be mailed at an oversize rate. What size package will maximize the volume while staying within the 108 inch limit?

X

Y

Z

Girth = 2x +2yVolume = x*y*z

10822

max0,0,0

zyxtosubject

xyzzyx

10822

max0,0,0

zyxtosubject

xyzzyx

First set up the lagrangian…

)22108(),,( zyxxyzzyx

Now, take derivatives…

0),,(

02),,(

02),,(

xyzyx

xzzyx

yzzyx

z

y

x

Lets assume lambda is positive 0

022108 zyx 022108 zyx

0

02

02

xy

xz

yz

xy

xz

xyyz

yz

2

02

02

yz

xyxz

xz

2

02

02

2

2

z x

z y

2 2 108

108

108

18

18

36

x y z

z z z

z

y

x

z

324xy

Suppose that you are able to produce output using capital (k) and labor (L) according to the following process:

5.5. lky

Labor costs $10 per hour and capital costs $40 per unit. You want to minimize the cost of producing 100 units of output.

klTC 4010

010

4010min

5..5

0,0

lktosubject

klkl

Minimizations need a minor adjustment…

yxgyxfyx ,,),(

A negative sign instead of a positive sign!!

)100(4010),( 5.5. lkllk

So, we set up the lagrangian again…now with a negative sign

Take derivatives…

05.40),(

05.10),(5.5.

5.5.

lklk

lklk

k

l

Lets again assume lambda is positive

05.40

05.105.5.

5.5.

lk

lk

01005.5. lk

5.5.

5.5.

20

05.10

kl

lk

5.5.

5.5.

80

05.40

lk

lk

kl

lkkl

4

8020 5.5.5.5.

200

50

1002

1004

1005.5.

5.5.

l

k

k

kk

lk

Suppose that you are choosing purchases of apples and bananas. Your total satisfaction as a function of your consumption of apples and bananas can be written as

6.4. BAU

Apples cost $4 each and bananas cost $5 each. You want to maximize your satisfaction given that you have $100 to spend

10054 BA

054100

max 6.4.

0,0

BAtosubject

BABA

First set up the lagrangian…

)54100(),( 6.4. BABABA

Now, take derivatives…

056.),(

044.),(4.4.

6.6.

BABA

BABA

B

A

054100

max 6.4.

0,0

BAtosubject

BABA

Lets again assume lambda is positive 0

AB

BABA

2.1

12.1. 4.4.6.6.

10054 BA

056.

044.4.4.

6.6.

BA

BA

6.1.

044.6.

6.6.

BA

BA

4.4.

4.4.

12.

056.

BA

BA

12

10

1002.154

10054

B

A

AA

BA

Suppose that you are able to produce output using capital (k) and labor (l) according to the following process:

5.5. lky

The prices of capital and labor are p and w respectively.Union agreements obligate you to use at least one unit of labor.

Assuming you need to produce y units of output, how wouldyou choose capital and labor to minimize costs?

1

min

5.5.

,

l

lky

tosubject

wlpklk

)1()(),( 5.5. lylkwlpklk

Just as in the previous problem, we set up the lagrangian. This time we have two constraints.

Doesn’t necessarily hold with equality

Will hold with equality

Non-Binding Constraints

.5 .5

.5 .5

.5 .5

( , ) .5 0

( , ) .5 0

0

1 0

k

l

k l p k l

k l w k l

y k l

l

First Order Necessary Conditions

01- 0 0)1( ll

)1()(),( 5.5. lylkwlpklk

0

05.

05.

5.5.

5.5.

5.5.

lky

lkw

lkp

First Order Necessary Conditions

Case #1: 1l

l

k

p

w

p

wll

p

wkly 2

0

1

w

pyl

2ypw

lp

wk

Constraint is non-binding

0

05.

05.

5.

5.

5.

ky

kw

kp

First Order Necessary Conditions

0Case #2: 1l

2yk

ypkp 22 22 ypypw

Constraint is binding

w

p

2ypw

w

pyl

p

wyk

Constraint is Non-Binding

Constraint is Binding

1

2

l

yk

Try this one…

You have the choice between buying apples and bananas. You utility (enjoyment) from eating apples and bananas can be written as:

BABAU 2, 5.

The prices of Apples and Bananas are given by BPAP and

Maximize your utility assuming that you have $100 available to spend

0 0

100$

2max 5.

,

BA

BPAP

tosubject

BA

BA

BA

BABPAPBABA BA 215. )100($2),(

(Income Constraint)

(Objective)

(You can’t eat negative apples/bananas!!)

ObjectiveNon-Negative Consumption Constraint

Income Constraint

BABPAPBABA BA 215. )100($2),(

0100$

02),(

05.),(

2

15.

BPAP

PBA

PABA

BA

BB

AA

First Order Necessary Conditions

We can eliminate some of the multiplier conditions with a little reasoning…

1. You will always spend all your income

2. You will always consume a positive amount of apples 01 0

0 0 0 22 BB

Case #1: Constraint is non-binding 0B 02

0100$

02),(

05.),( 5.

BPAP

PBA

PABA

BA

BB

AA

First Order Necessary Conditions

BA PP

A 25. 5.

B

A

P

APB

100$ AB PPB 400

Case #1: Constraint is binding 0B 02

0100$

02),(

05.),(

2

5.

AP

PBA

PABA

A

BB

AA

First Order Necessary Conditions

AA PA

P

A 100$

5. 5.

22 BPAB PP 4002

BP

AP

AB PP 40

Constraint is Non-Binding

Constraint is Binding

APA

B

100$

0

B

A

P

APB

100$2

25.

A

B

P

PA