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)
B
f :A B
f
Is the range
is a function
A
f
A
B
Is the domain
For example
For
0 x5
Domain
f  x   3x  5
Range
Function
A  0,5
3
x A
5  y  20
B  [5,20]
f 3  14
14
y
y  3x  5
20
Y =14
0 x5
Range
For
5
0
Domain
X =3
x
5
max 3x  5
y
20
0 x5
Range
Here, the optimum occurs at
x = 5 (y = 20)
5
0
Domain
x
5
Optimization involves finding the maximum value for y over an allowable
domain.
What is the solution to this optimization problem?
 1 
max 

5  x 
y
0  x  10
5
10
x
There is no optimum because
f(x) is discontinuous at x = 5
What is the solution to this optimization problem?
max 2 x
y
0 x6
12
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
0
6
What is the solution to this optimization problem?
y
max
 x
x0
12
There is no optimum because
the domain is unbounded (x is
allowed to become arbitrarily
large)
x
0
The Weierstrass theorem provides sufficient conditions for an optimum
to exist, the conditions are as follows:
y
f x  is continuous
max  f ( x)
over the domain of
x
The domain for
x
is closed and bounded
x
Finding maxima/minima involves taking derivatives….
Formally, the derivative of
f x 
is defined as follows:
 f ( x  x)  f ( x) 
f '  x   lim 

x 0
x


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
Graphically…
y
y
f x  x
f x 
f x 
x
x  x
f ( x  x)  f ( x)
slope 
x
x
x
x
Now, let the change
in x get arbitrarily
small
 f ( x  x)  f ( x) 
f '  x   lim 

x 0
x


First Order Necessary Conditions
If
x* is a solution to the optimization problem
max  f ( x)
min  f ( x )
or
x
x
then
f ' ( x*)  0
f x 
f ' x   0
f ' x   0
x*
f x 
f ' x   0
x
f ' x   0
f ' x   0
x*
x
f ' x   0
Useful derivatives
Logarithms
Linear Functions
f ( x)  Ax  f ' ( x)  A
Example:
f ( x)  4 x
f ' x   4
A
f ( x)  A ln( x)  f ' ( x) 
x
Example:
Exponents
f ( x)  Ax n  f ' ( x)  nAx n1
Example:
f ( x)  3x 5
f ' x   15 x 4
f ( x)  12 ln  x 
12
f ' x  
x
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
$750,000
Hourly Cost 
 $1500  $1.5 x
x
 $750,000

min 
 $1500  $1.5 x 
x 0
x


First Order Necessary Condition
 $750,000
750,000
 $1.5  0  x 
 707hrs
2
x
1.5
Hourly Cost ($)
An Example
Annual Flight Hours
How can we be sure we are at a maximum/minimum?
For a maximization…
For a minimization…
f x 
f x 
f ' x   0
f ' x   0
f ' x   0
x*
x
f ' x   0
f ' x   0
x*
x
f ' x   0
f ' ' ( x*)  0
f ' ' ( x*)  0
Slope is decreasing
Slope is increasing
Let x = Number of Flight Hours
 $750,000

min 
 $1500  $1.5 x 
x
x


First Order Necessary Conditions
 $750,000
750,000
 $1.5  0  x 
 707hrs
2
x
1.5
Second Order Necessary Conditions
$1,500,000
0
3
x
For X>0
Suppose you know that demand for your product depends on the
price that you set and the level of advertising expenditures.
Q p, A  5,000  10 p  40 A  pA  .8 A2  .5 p 2
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
max 5,000  10 p  40 A  pA  .8 A2  .5 p 2 
p  0 , A 0
First Order Necessary Conditions
Q p ( p, A)  10  A  p  0
QA ( p, A)  40  p 1.6 A  0
With our two first order
conditions, we have two
variables and two unknowns
 10  A  p  0
A  p  10
Q p ( p, A)  10  A  p  0
QA ( p, A)  40  p 1.6 A  0
40  p  1.6 A  0
40  p  1.6 p  10   0
24  .6 p  0
p  40
A  50
How can we be sure we are at a maximum?
f ' ' ( x*)  0
its generally sufficient to see if all the second derivatives are negative…
Q p ( p, A)  10  A  p
QA ( p, A)  40  p 1.6 A
Q pp ( p, A)  1  0
QAA ( p, A)  1.6  0
Practice Questions
1) Suppose that profits are a function of quantity produced and can be
written as
  10  20Q  2Q
2
Find the quantity that maximizes profits
2) Suppose that costs are a function of two inputs and can be written as
C  4  3 X1  4 X 2  2 X  3 X  X1 X 2
2
1
2
2
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
max f  x, y 
x, y
subject to
g  x,y  0
To solve these types of problems, we set up the lagrangian
( x, y)  f x, y   g x, y 
Function to be
maximized
Constraint(s)
Multiplier
To solve these types of problems, we set up the lagrangian
( x)  f x  g x 
0
(x ) f x 
We know that at the
maximum…
(x )
f x 
x*
x
 x ( x)  0
g x  0
Once you have set up the lagrangian, take the derivatives and set
them equal to zero
( x, y)  f x, y   g x, y 
First Order Necessary Conditions
 x ( x, y )  f x  x, y    g x  x, y   0
 y ( x , y )  f y  x , y    g y  x, y   0
Now, we have the “Multiplier” conditions…
 0
g x, y   0
g x, y   0
Example: Suppose you sell two products ( X and Y ). Your profits as a
function of sales of X and Y are as follows:
 ( x, y)  10 x  20 y  .1( x 2  y 2 )
Your production capacity is equal to 100 total units. Choose X and Y
to maximize profits subject to your capacity constraints.
x  y  100
The key is to get
the problem in the
right format
f x, y 
max 10 x  20 y  .1( x 2  y 2 )
x 0, y 0
subject to
100  x  y  0
g x, y 
Multiplier
The first step is to create a Lagrangian
( x, y)  10 x  20 y  .1( x  y )   (100  x  y)
2
Objective Function
2
Constraint
Now, take the derivative with respect to x and y
( x, y)  10 x  20 y  .1( x  y )   (100  x  y)
2
2
First Order Necessary Conditions
 x ( x, y )  10  .2 x    0
 y ( x, y )  20  .2 y    0
“Multiplier” conditions
 0
100  x  y  0
 (100  x  y )  0
First, lets consider the possibility that
lambda equals zero
10  .2 x    0
20  .2 y    0
10  .2 x  0
x  50
 0
100  x  y  0
 100  x  y   0
20  .2 y  0
y  100
100  50 100  50  0
Nope! This can’t work!
The other possibility is that lambda is
positive
10  .2 x    0
20  .2 y    0
  10  .2x
  20  .2 y
10  .2 x  20  .2 y
y  x  50
y  x  50
 0
 100  x  y   0
100  x  y  0
100  x  y  0
100  x   x  50   0
50  2 x  0
x  25
y  75
Lambda indicates the marginal value of relaxing the constraint. In this
case, suppose that our capacity increased to 101 units of total
production.
  10  .2 x  20  .2 y
y  75
x  25
 50
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?
Z
Girth = 2x +2y
Volume = x*y*z
X
Y

xyz
x 0, y 0, z 0
max
subject to
2 x  2 y  z  108

xyz
x 0, y 0, z 0
max
subject to
2 x  2 y  z  108
First set up the lagrangian…
( x, y, z )  xyz   (108  2 x  2 y  z )
Now, take derivatives…
 x ( x, y, z )  yz  2  0
 y ( x, y, z )  xz  2  0
 z ( x, y, z )  xy    0
Lets assume lambda is positive
yz  2  0
xz  2  0
xy    0
108  2 x  2 y  z  0
  xy
yz  2  0
yz  2 xy  0
z  2x
xz  2  0
xz  2 xy  0
z  2y
 0
 108  2x  2 y  z   0
z  2x
z  2y
2 x  2 y  z  108
z  z  z  108
z  108
y  18
  xy  324
x  18
z  36
Suppose that you are able to produce output using capital (k) and labor (L) according
to the following process:
yk l
.5 .5
Labor costs $10 per hour and capital costs $40 per unit. You want to minimize the
cost of producing 100 units of output.
TC  10l  40k
min 10l  40k
l 0,k 0
subject to
k .5l .5  100
Minimizations need a minor adjustment…
( x, y)  f x, y   g x, y 
A negative sign instead of a positive sign!!
So, we set up the lagrangian again…now with a negative sign
(k , l )  10l  40   (k .5l .5  100)
Take derivatives…
 l (k , l )  10  .5k l
.5 .5
0
 k (k , l )  40  .5k .5l .5  0
Lets again assume lambda is positive
 0
k l  100
.5 .5
10  .5k l
.5 .5
0
40  .5k l  0
.5 .5
40  .5k .5l .5  0
  80k .5l .5
10  .5k .5l .5  0
  20l .5 k .5
k .5l .5  100
.5
20l k
l  4k
.5
.5 .5
 80k l
k .5 4k   100
2k  100
.5
k  50
l  200
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
UA B
.4
.6
Apples cost $4 each and bananas cost $5 each. You want to maximize your
satisfaction given that you have $100 to spend
4 A  5B  100
.4
max A B
.6
A 0 , B  0
subject to 100  4 A  5B  0
max A.4 B.6
A 0 , B  0
subject to 100  4 A  5B  0
First set up the lagrangian…
( A, B)  A B   (100  4 A  5B)
.4
.6
Now, take derivatives…
 A ( A, B)  .4 A.6 B.6  4  0
 B ( A, B)  .6 A.4 B .4  5  0
Lets again assume lambda is positive
.4 A B  4  0
.6
.4
.6 A B
.6
.4
 0
4 A  5B  100
 5  0
.6 A.4 B .4  5  0
  .12 A.4 B .4
.4 A.6 B.6  4  0
  .1A.6 B.6
.1A.6 B .6  .12 A.4 B .4
B  1.2 A
4 A  5B  100
4 A  51.2 A  100
A  10
B  12
Suppose that you are able to produce output using capital (k) and
labor (l) according to the following process:
yk l
.5 .5
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 would
you choose capital and labor to minimize costs?
Non-Binding Constraints
min pk  wl
k ,l
subject to
y  k .5l .5
l 1
Just as in the previous problem, we
set up the lagrangian. This time we
have two constraints.
Will hold with equality
(k , l )  pk  wl   (k .5l .5  y )   (l  1)
Doesn’t necessarily hold with equality
(k , l )  pk  wl   (k .5l .5  y )   (l  1)
First Order Necessary Conditions
.5 .5
(
k
,
l
)

p


.5
k
 l 0
k
(k , l )  w   .5k l
.5 .5
l
  0
y  k .5l .5  0
l 1  0
 (l  1)  0
 0
l -1  0
Case #1:
 0
l 1
Constraint is non-binding
First Order Necessary Conditions

w   .5k

 0
p   . 5 k  .5 l . 5  0
.5  . 5
l
w k

p l
 w
k   l
 p
y  k .5 l . 5  0
 w 2
y  kl   l  l
 p
 w
 
 p
 p
l  y   1
 w
w  py 2
Case #2:
l 1
 0
Constraint is binding
First Order Necessary Conditions
 
w   .5k     0
p   .5k .5  0
.5
yk 0
.5
  2 p k  2 py
ky
2
w  py 2    py 2
w
Constraint is Binding
w  py 2
k  y2
l 1
Constraint is Non-Binding
 p
ly  
 w
 w
k  y  
 p
p
Try this one…
You have the choice between buying apples and bananas. You utility
(enjoyment) from eating apples and bananas can be written as:
U  A, B   A  2 B
.5
The prices of Apples and Bananas are given by
PA
and
PB
Maximize your utility assuming that you have $100 available to
spend

max A  2 B
A, B
.5

(Objective)
subject to
PA A  PB B  $100
A0 B0
(Income Constraint)
(You can’t eat negative apples/bananas!!)
( A, B)  A.5  2B   ($100  PA A  PB B)  1 A  2 B
Objective
Income
Constraint
Non-Negative
Consumption
Constraint
( A, B)  A.5  2B   ($100  PA A  PB B)  1 A  2 B
First Order Necessary Conditions
 A ( A, B )  .5 A.5  PA  1  0
 B ( A, B )  2  PB   2  0
$100  PA A  PB B  0
We can eliminate some of the multiplier conditions with a little reasoning…
1. You will always spend all your income
 0
2. You will always consume a positive amount of apples
2  0
B0
2 B  0
1  0
Case #1: Constraint is non-binding
B  0 2  0
First Order Necessary Conditions
 A ( A, B )  .5 A.5  PA  0
 B ( A, B )  2  PB  0
$100  PA A  PB B  0
.5
.5 A
PA
2
 
PB
$100  PA A
B
PB
B  0  PB  40 PA
Case #1: Constraint is binding
B  0 2  0
First Order Necessary Conditions
 A ( A, B)  .5 A.5  PA  0
 B ( A, B)  2  PB   2  0
$100  PA A  0
.5 A.5

PA
$100
A
PA
2  PB  2
2  0  PB  40 PA
PB
Constraint is Binding
B0
PB  40 PA
$100
A
PA
Constraint is Non-Binding

PB 
A   .25 
PA 

2
$100  PA A
B
PB
PA