optimization introduction

1. FIN 30210:
Managerial Economics
Optimization
Techniques

2. Economics is filled with optimization problems
Consumers make buying
decisions to maximize utility
Consumers make labor
decisions to maximize utility
Businesses make hiring/capital
investment decisions to
minimize costs Businesses make pricing
decisions to maximize profits
Consumers make savings
decisions to maximize utility
We need some tools
to analyze these
optimization
problems

3. Example An airport shuttle currently charges $15 and
carries an average of 1,200 passengers per day. It
estimates that for each dollar it raises its fare, it
loses an average of 50 passengers.
Current Revenues: $15(1,200) = $18,000
Could we do better?

4. Passengers
Price
First, lets figure out the demand curve this business faces. For simplicity, lets assume
that the demand is linear.
1,200
$15
Q A BP
 
A
A
B
*0
Q A B A
  
0 A BP
 
A
P
B

We also know that every dollar change in price alters passengers
by 50.
50
B 

5. Passengers
Price
First, lets figure out the demand curve this business faces. For simplicity, lets assume
that the demand is linear.
1,200
$15
 
1,200 50 15
A
 
1,950
1,950
$39
50

We can use the one data point we have to find the
missing parameter
1,950
A  1,950 50
Q P
 

6. Passengers
Fare
Q
$P
1,950
1,950
$39
50

The problem here is to maximize revenues.
Revenues *
P Q

Revenues
  2
Revenues 1,950 50 1,950 50
P P P P
   
1,950 50
Q P
 
So, we have revenues as a
function of the price charged.
Now what?

7. Price Quantity Revenues
0 1,950 0
1 1,900 1,900
2 1,850 3,700
3 1,800 5,400
1,950 50
Q P
  2
1,950 50
P P

Price Quantity Revenues
36 150 5,400
37 100 3,700
38 50 1,900
39 0 0
0
2000
4000
6000
8000
10000
12000
14000
16000
18000
20000
1 5 9 13 17 21 25 29 33 37
15
19.50
Price Quantity Revenues
19.50 975 19,012.50
We could do better! Now, how do we find this
point without resorting to excel?
Price
Revenues

8. “Take the derivative and set it equal
to zero!”
-Fermat’s Theorem
Pierre de Fermat
1607- 1665 Pierre de Fermat was actually a lawyer before he
was a mathematician….perhaps he was trying to
figure out how to maximize his legal fees!
But, what’s a derivative and why set it
equal to zero?

9. x
 
x
f
x
 
x
f
x x
 
 
f x x
 
   
f x x f x
slope
x
  


x
 
x
f
x
 
x
f
   
 
'
0
lim
x
f x x f x
slope f x
x
 
  
 
 
 

 
 
'
f x
Imagine calculating the slope
between two distinct points on a
function
Now, let those two points get
closer and closer to each
other
The derivative is the result of the
distance between those two
points approaching zero

10. x
 
x
f
2
4
6
Slope
4 36 8
2 16 6
1 9 5
.5 6.25 4.5
.25 5.0625 4.25
.1 4.41 4.1
.05 4.2025 4.05
.01 4.0401 4.01
.001 4.004001 4.001
2
x
x
  
f x x
 
   
f x x f x
slope
x
  


2
36
36 4
8
4
slope

 
x

Let’s try one numerically…
The slope gets closer and closer to 4

11. x
 
x
f
x
 
f x
x x
 
2
x
 
f x x
 
x

Or, in general…    
f x x f x
slope
x
  


 
2 2
x x x
slope
x
  


2 2 2
2
x x x x x
slope
x
    


2
2x x x
slope
x
  


2
slope x x
  
Let the change in x go to 0
2
slope x

 
'
2 4
f 

12. Some useful derivatives
Linear Functions
A
x
f
Ax
x
f 

 )
(
'
)
(
Example:
  4
'
4
)
(


x
f
x
x
f
x
A
x
f
x
A
x
f 

 )
(
'
)
ln(
)
(
Logarithms
Example:  
 
x
x
f
x
x
f
12
'
ln
12
)
(


Exponents
1
)
(
'
)
( 


 n
n
nAx
x
f
Ax
x
f
Example:
  4
5
15
'
3
)
(
x
x
f
x
x
f



13. x
 
x
f
*
x
 
'
0
f x 
x
 
x
f
*
x
 
'
0
f x 
A necessary condition for a maximum or a minimum is that the
derivative equals zero
OR
But how can we tell which is which?

14. x
 
x
f
*
x
 
'
0
f x 
 
'
0
f x 
As x increases, the
slope is decreasing
 
''
0
f x 
x
 
x
f
*
x
 
'
0
f x 
As x increases, the
slope is increasing
 
''
0
f x 
 
'
0
f x 
While the first derivative measures the slope (change in the value of the function), the
second derivative measures the change in the first derivative (change in the slope)

15. William Sears: “of course we were embarrassed by the error”
The B2 Bomber: A lesson in second derivatives
“The flying wing was the aerodynamically worst
possible choice of configuration”
--Joseph Foa

16. 0
2000
4000
6000
8000
10000
12000
14000
16000
18000
20000
1 5 9 13 17 21 25 29 33 37
15
19.50
Back to our revenue problem….
2
1,950 50
R P P
 
Take the
derivative with
respect to ‘P’
'
1,950 100
R P
 
Set the derivative
equal to zero and
solve for ‘P’
1,950 100 0
P
 
$19.50
$19,012.50
P
R


Note, the second derivative is
negative…a maximum!
''
100
R  
Revenues
Price

17. After prices for almonds climbed to a record $4 per pound in 2014, farmers
across California began replacing their cheaper crops with the nut, causing
a huge increase in supply. Now, the bubble has popped. Since late 2014,
according to The Washington Post, almond prices have fallen by around
25%.
Lets suppose that almond prices
are currently $3/lb. and are falling
at the rate of $.04/lb. per week.
Let’s also suppose that your
orchard is current bearing 35
pounds per tree, and that number
will increase by 1 lb. per week.
How long should you wait to
harvest your nuts to maximize your
revenues?
$3.00

18. 35
Q t
  3.00 .04
P t
 
We want to maximize revenues which is price
per pound times total pounds sold
  
Revenue * 35 3 .04
P Q t t
   
2
105 1.6 .04
R t t
  
Now, take a
derivative with
respect to ‘t’ and set
it equal to zero
Then, solve for ‘t’
' 1.6 .08
R t
 
*
20
t 
$2.20
55
$121
P
Q
R




19. 95
100
105
110
115
120
125
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27
We want to maximize revenues which is price per pound times total pounds sold
Weeks
Revenues

20. Suppose that you run a
trucking company. You have
the following expenses.
Expenses
• Driver Salary: $22.50 per hour
• $0.27 per mile depreciation
• v/140 dollars per mile for fuel
costs where ‘v’ is speed in miles
per hour
What speed should you
tell your driver to maintain
in order to minimize your
cost per mile?

21. Note that if I divide the
driver’s salary by the speed, I
get the driver’s salary per
mile
$
$
.
.
hr
miles mile
hr

22.50
.27
140
v
Cost
v
  
What speed should you tell your driver to
maintain in order to minimize your cost per
mile?
Now, take a
derivative with
respect to ‘t’ and set
it equal to zero
Then, solve for ‘t’
2
22.50 1
' 0
140
C
v
   
3,150 56
$1.07
v mph
Cost per mile
 


22. 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72
1.05
1.06
1.07
1.08
1.09
1.1
We want to minimize cost per mile
Speed
Cost
Per
Mile

23. Suppose you know that demand for your
product depends on the price that you set and
the level of advertising expenditures.
  2
2
5
.
8
.
40
10
000
,
5
, p
A
pA
A
p
A
p
Q 





Choose the level of advertising AND price top maximize
sales

24. 2 2
5,000 10 40 .8 .5
Q p A pA A p
     
Choose the level of advertising AND price to maximize sales
Now, we need
partial derivatives
with respect to both
‘p’ and ‘A’. Both are
set equal to zero
10 0
p
Q A p
    
40 1.6 0
A
Q p A
   
40 1.6 0
p A
  
10 0
A p
   
This gives us two
equations with two
unknowns

25. 10 0 10
A p A p
      
40 1.6 0
p A
  
Advertising
Price
25
-40
40 1.6 0
p A
  
-10
10
10 0
A p
   
 
40 1.6 0
40 1.6 10 0
24 .6 0
40
50
p A
p p
p
p
A
  
   
 


50
40
We could solve one of the equations
for ‘A’ and then plug into the other

26. Joseph-Louis Lagrange
1736-1813
The method of Lagrange multipliers is a strategy for
finding the local maxima and minima of a function
subject to equality constraints.
Harold W. Kuhn
1925 - 2014
Albert W. Tucker
1905 - 1995
Harold Kuhn and Albert Tucker later
extended this method to inequality
constraints

27.  
  0
,
max
,

x,y
g
to
subject
y
x
f
y
x
The Lagrange method writes the constrained optimization
problem in the following form
Constraint(s)
Objective
Choice
variables
The problem is then rewritten as follows
   
, ,
f x y g x y

 
Multiplier (assumed greater or equal to zero)

28.    
f x g x

 
 
f x
  0
g x
 
Here’ s how this would look in two dimensions…we’ve created a new function that includes
the constraints. This new function coincides with the original objective at one point – the
maximum of both!
x
 
f x
*
x
Therefore, at the maximum,
two thigs have to be true
0
x 
  0

x
g

The new function is
maximized
The new function coincides
with the original objective

29.    
, ,
f x y g x y

 
So, we have our Lagrangian function….
We need the derivatives with respect o both ‘x’ and ‘y’ to be zero
   
   
, , 0
, , 0
x x x
y y y
f x y g x y
f x y g x y


  
  
And then we have the “multiplier conditions”
    0
,
0
,
0 

 y
x
g
y
x
g 


30. Example: Suppose you sell two products ( X and Y ). Your profits
as a function of sales of X and Y are as follows:
2 2
Profit 10 20 .1( )
x y x y
   
Objective
 
,
f x y
100

 y
x
Your production capacity is equal to 100 total units. Choose X
and Y to maximize profits subject to your capacity constraints.
Constraint
 
,
g x y

31. First, for comparison purposes, lets solve this unconstrained….
2 2
Profit 10 20 .1( )
x y x y
   
Take derivatives with respect to ‘x’ and ‘y’
P 10 .2
x x
 
P 20 .2
y y
 
Set the derivatives equal to
zero and solve for ‘x’ and ‘y’
*
*
50
100
1,250
x
y
P



50 x
y
100
1,250
P 

32. Now, add the constraint 2 2
Profit 10 20 .1( )
x y x y
   
50
x
y
100
1,250
P 
100

 y
x
Acceptable
values for ‘x’
and ‘y’
Subject to
The constraint matters!
We need to get the problem in the right format….we need
unconstrained
solution  
, 0
g x y  100

 y
x
Subtract ‘x’ and ‘y’
from both sides
100 0
x y
  
Now, we can write the lagrangian
2 2
10 20 .1( ) (100 )
x y x y x y

      
Objective Constraint
Multiplier
100

33. Now, the mechanical
part
50
x
y
100
1,250
P 
Acceptable
values for ‘x’
and ‘y’
unconstrained
solution
2 2
10 20 .1( ) (100 )
x y x y x y

      
10 .2 0
20 .2 0
x
y
x
y


   
   
Take derivatives with respect to ‘x’ and ‘y’
And the multiplier conditions
0
)
100
(
0
100
0 





 y
x
y
x 


34. Now, the mechanical
part
50
x
y
100
1,250
P 
Acceptable
values for ‘x’
and ‘y’
unconstrained
solution
10 .2 0
20 .2 0
x
y
x
y


   
   
0
)
100
(
0
100
0 





 y
x
y
x 

First, let’s suppose that lambda equals zero
10 .2 0
20 .2 0
x
y
 
 
100 0
x y
  
*
*
50
100
x
y


100 50 100 50 0
    
Nope, that doesn’t work!! So, lambda
must be a positive number
That puts us here!

35. Now, the mechanical
part
50
x
y
100
1,250
P 
unconstrained
solution
10 .2 0
20 .2 0
x
y
x
y


   
   
0 (100 ) 0 100 0
x y x y
 
      
So, we have three equations and three unknowns (‘x’, ‘y’, and lambda)
10 .2 0
20 .2 0
x
y


  
  
10 .2
20 .2
x
y


 
 
Solve the first
two expressions
for lambda
10 .2 20 .2
x y
  
50
y x
 
rearrange
 
100 50 0
x x
   
*
*
25
75
5
1,125
x
y
P





Plug into third equation
25
75
1,125
P 
constrained solution

36. Suppose that the
production constraint
increases from 100 to 101
50
x
y
100
1,250
P 
unconstrained
solution
25
75
1,125
P  (x + y = 100)
10 .2 0
20 .2 0
x
y
x
y


   
   
0 (101 ) 0 101 0
x y x y
 
      
Resolve….
 
101 50 0
x x
   
*
*
25.5
75.5
4.9
1,,129.5
x
y
P





1,129.5
P  (x + y = 101)
We optimize once again and, given the extra capacity,
profits up by 4.50. This is pretty close to the value of
lambda!

37. #
x y
 
150
#
1,250
P 
100
unconstrained
solution
Profit
Let’s plot profits (optimized subject to the constraint) as a function of
the constraint
Slope 

1,125
P 
Lambda measures the marginal impact of
the constraint on the objective function!

38. Suppose that the production
constraint is 160
50
x
y
100
1,250
P 
unconstrained
solution
10 .2 0
20 .2 0
x
y
x
y


   
   
0 (160 ) 0 160 0
x y x y
 
      
*
*
105
155
1
1,245
x
y
P



 

160
160
If we continue to assume lambda is positive….
 
160 50 0
x x
   
This is no good! Lambda must
be zero then!
*
*
50
100
0
1,250
x
y
P





1,245
P 

39. #
x y
 
150
#
1,250
P 
unconstrained
solution
Profit Slope 

Lambda measures the marginal impact of the constraint on the objective
function! So, when the constraint in irrelevant, lambda hits zero!
Profit
Lambda

40. 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?
Y Z
X
 
108
2
2
max
0
,
0
,
0






z
y
x
to
subject
xyz
z
y
x
2 2
Girth x y
Volume xyz
 

 
, , 0
g x y z  108 2 2 0
x y z
   
Remember, we need to write the
constraint in the right format!
Example

41. Y Z
X
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?
(108 2 2 )
xyz x y z

    
First, write down the lagrangian
2 0
2 0
0
x
y
z
yz
xz
xy



  
  
  
Set the derivatives equal to zero
  0
2
2
108 


 z
y
x

0
  108 2 2 0
x y z
   
Multiplier conditions
Now, take the derivatives with respect to ‘x’, ‘y’, and ‘z’

42. Lets assume that lambda is positive
That leaves us four equations and 4 unknowns
2 0
2 0
0
yz
xz
xy



 
 
 
108 2 2 0
x y z
   
x
z
xy
yz
yz
2
0
2
0
2




 
2 0
2 0
2
xz
xz xy
z y

 
 

xy


2 2 108
108
3 108
18
18
36
x y z
z z z
z
y
x
z
  
  




   
18 18 36 11,664
Area xyz
  
  
18 18 324
xy
   
Adding 1 inch to the girth/length will allow us to
increase the area by approximately 324 cubic inches

43. Example
5
.
5
.
l
k
y 
Suppose that you manufacture IPods. Your assembly
plant utilizes both labor and capital inputs. You can
write your production process as follows
Labor costs $10 per hour and capital costs $40 per unit. Your objective is to
minimize the production costs associated with producing 100 IPods per hour.
10 40
Total Costs l k
 
Hourly output of
IPods Capital Inputs
Labor inputs

44. 0
10
40
10
min
5
.
.5
0
,
0




l
k
to
subject
k
l
k
l
Minimize the production costs associated with
producing 100 IPods per hour.
 
, , 0
g x y z  .5 .5
100 0
k l  
Remember, we need to write the
constraint in the right format!
Minimizations need a minor adjustment…
   
, ,
f x y g x y

 
A negative sign instead of a positive sign!!

45. k
l
.5 .5
100 k l

16
625
   
.5 .5
16 625 100

100
100
   
.5 .5
100 100 100

200
50
   
.5 .5
200 50 100

0
10
40
10
min
5
.
.5
0
,
0




l
k
to
subject
k
l
k
l
The shaded area is what the constraint
look like

46. Minimize the production costs associated with
producing 100 IPods per hour.
.5 .5
10 40 ( 100)
l k l

   
So, we set up the lagrangian again…now with a negative sign
Now, take the derivatives with respect to ‘k’ and ‘l’
.5 .5
.5 .5
10 .5 0
40 .5 0
l
k
k l
k l




  
  
Set the derivatives equal to zero
Multiplier conditions
0
  .5 .5
100
k l 
.5 .5
( 100) 0
k l
  

47. Lets assume that lambda is positive. So we have three equations and three unknowns
.5 .5
.5 .5
.5 .5
10 .5 0
40 .5 0
100
k l
k l
k l




 
 
 5
.
5
.
5
.
5
.
20
0
5
.
10





k
l
l
k


Lets solve the first two for lambda
Set the two
expressions for
lambda equal to each
other and simplify
k
l
l
k
k
l
4
80
20 5
.
5
.
5
.
5
.

 

Plug into the last
expression and
simplify
 
200
50
100
2
100
4
100
5
.
5
.
5
.
5
.





l
k
k
k
k
l
k
   
10 200 40 50 $4,000
Total Costs   
   
.5 .5
.5 .5
20 20 200 50 40
l k



  
5
.
5
.
5
.
5
.
80
0
5
.
40





l
k
l
k


1 Additional
IPod would cost
$40 (wait, isn’t
that marginal
cost?!)

48. k
l
.5 .5
100 k l

16
625
100
100
200
50
0
10
40
10
min
5
.
.5
0
,
0




l
k
to
subject
k
l
k
l
The shaded area is what the constraint
look like
50
200
   
10 50 40 200 8,500
Total Costs   
   
10 100 40 100 5,000
Total Costs   
   
10 200 40 50 4,000
Total Costs   
   
10 625 40 16 6,890
Total Costs   

49. 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
.4 .6
Utility A B

Happiness
Weekly Apple
Consumption
Weekly Banana
Consumption
Apples cost $4 each and bananas cost $5 each. You want to maximize your
satisfaction given that you have $100 to spend
4 5
Total Expenditures A B
 

50. B
A
$100
25
$4

$100
20
$5

4 5 100
Total Expenditures A B
  
4 5 100
A B
 
.4 .6
0, 0
max
100 4 5 0
A B
A B
subject to A B
 
  
*Note: Technically, there are two
additional constraints here
0 0
A B
 
But given that zero consumption of
either apples or oranges yields zero
utility, we can ignore them

51. Apples cost $4 each and bananas cost $5 each. You want to
maximize your satisfaction given that you have $100 to spend
.4 .6
0, 0
max
100 4 5 0
A B
A B
subject to A B
 
  
)
5
4
100
(
)
,
( 6
.
4
.
B
A
B
A
B
A 


 

So, we set up the lagrangian again
Take derivatives with respect to ‘A’ and
‘B’ and set the derivatives equal to zero
.6 .6
.4 .4
.4 4 0
.6 5 0
A
B
A B
A B




  
  
Let’s again assume
lambda is positive!
100 4 5 0
A B
  
0
 

52. So we have three equations and three unknowns
.6 .6
.4 .4
.4 4 0
.6 5 0
4 5 100
A B
A B
A B




 
 
 
Lets solve the first two for lambda
.6 .6
.6 .6
.4 4 0
.1
A B
A B




 
 4
.
4
.
4
.
4
.
12
.
0
5
6
.





B
A
B
A


Set the two
expressions for
lambda equal to each
other and simplify
A
B
B
A
B
A
2
.
1
12
.
1
. 4
.
4
.
6
.
6
.

 

Plug into the last
expression and
simplify
 
12
10
100
2
.
1
5
4
100
5
4






B
A
A
A
B
A
   
.6 .6
.6 .6
.1 .1 10 12 .11
A B



  
   
.4 .6
.4 .6
10 12 11.15
Utility A B
  
1 Additional
dollar would
provide an
additional .11
units of
happiness

53. B
A
$100
25
$4

$100
20
$5

4 5 100
Total Expenditures A B
  
0
5
4
100
max 6
.
4
.
0
,
0





B
A
to
subject
B
A
B
A
   
.4 .6
.4 .6
10 12 11.15
Utility A B
  
10
12

54. Non-Binding Constraints

55. k
l
TC 40
10 

Suppose that you manufacture IPods. Your assembly plant utilizes both
labor and capital inputs. You can write your production process as follows
5
.
5
.
l
k
y 
Hourly output of
IPods Capital Inputs
Labor inputs
Labor costs $10 per hour and capital costs $40 per unit. Your objective is to minimize the
production costs associated with producing 100 IPods per hour.
The local government will not allow you to fully automate your plant. That is, you have
to use at least 1 hour in your production process
1
l 

56. k
l
.5 .5
100 k l

0
10
40
10
min
5
.
.5
0
,
0




l
k
to
subject
k
l
k
l
1
l 
1
1
l 
Possible
Choices

57. Minimize the production costs associated with
producing 100 IPods per hour while utilizing at least 1
hour of labor.
 
.5 .5
10 40 ( 100) 1
l k k l l
 
     
Labor constraint
Take derivatives with respect to ‘k’ and
‘l’ and set the derivatives equal to zero
 
 
.5 .5
.5 .5
40 .5 0
10 .5 0
k
l
k l
k l

 


  
   
0
1
-
0
0
)
1
( 


 l
l 

 
.5 .5 .5 .5
0 100 0 100 = 0
k l k l
 
   
Multiplier Conditions

58.  
 
.5 .5
.5 .5
40 .5 0
10 .5 0
k l
k l

 


 
  
We already know that that lambda is positive, but what about the other multiplier?.
Suppose that mu is positive as well (i.e. the second constraint is binding)
.5 .5
100 0
k l  
1
l 
So, we can plug in labor equals 1
 
 
.5
.5
40 .5 0
10 .5 0
k
k

 

 
  
.5
100 0
k   10,000
k 
 
 
.5
.5
40 .5 0
10 .5 0
k
k

 

 
  
Plug in capital equals 10 and solve for lambda and mu
This doesn’t work. That means the second constraint is not binding and so we can ignore it
8,000
 
399,990
  

59. k
l
.5 .5
100 k l

0
10
40
10
min
5
.
.5
0
,
0




l
k
to
subject
k
l
k
l
1
l 
1
1
l 
50
200
   
10 200 40 50 4,000
Total Costs   

60. Lets make sure the constraint binds… suppose that
labor costs $500,000 per hour!!
 
.5 .5
500,000 40 ( 100) 1
l k k l l
 
     
Labor constraint
Take derivatives with respect to ‘k’ and
‘l’ and set the derivatives equal to zero
 
 
.5 .5
.5 .5
40 .5 0
500,000 .5 0
k
l
k l
k l

 


  
   
0
1
-
0
0
)
1
( 


 l
l 

 
.5 .5 .5 .5
0 100 0 100 = 0
k l k l
 
   
Multiplier Conditions

61.  
 
.5 .5
.5 .5
40 .5 0
500,000 .5 0
k l
k l

 


 
  
We already know that that lambda is positive, but what about the other multiplier?.
Suppose that mu is positive as well (i.e. the second constraint is binding)
.5 .5
100 0
k l  
1
l 
So, we can plug in labor equals 1
 
 
.5
.5
40 .5 0
500,000 .5 0
k
k

 

 
  
.5
100 0
k   10,000
k 
 
 
.5
.5
40 .5 0
500,000 .5 0
k
k

 

 
  
Plug in capital equals 10 and solve for lambda and mu
8,000
 
100,000
 
Now the constraint binds!

62. k
l
.5 .5
100 k l

1
l 
1
1
l 
10,000
   
500,000 1 40 10,000 900,000
Total Costs   
0, 0
.5 .5
min 500,000 40
100
l k
l k
subject to k l
 



63. 1
l 
0, 0
.5 .5
min
l k
wl pk
subject to k l y
 


So, when does the constraint bind? Lets solve the problem for a generic price of capital, wage,
and production requirement
 
 
0
0
5
.
0
5
.
5
.
5
.
5
.






 
k
y
k
w
k
p



2
y
k 
y
p
k
p 2
2 

 2
2
y
p
y
p
w 

 
Assuming L = 1

64. Hourly
Wage
w
p Price of
capital
per unit
2
y
p
w 
1
2


l
y
k







w
p
y
l 








p
w
y
k
Constraint is Non-Binding
Constraint is Binding
Whether or not the constraint binds depends on prices!!

65. Let’s go back to the banana/apple problem. However,
lets change up the utility function.
 
0
0
100
$
2
max 5
.
,





B
A
B
P
A
P
to
subject
B
A
B
A
B
A
With the new utility function, we can no longer assume that
positive amounts of apples and bananas are consumed!

66.  
0
0
100
$
2
max 5
.
,





B
A
B
P
A
P
to
subject
B
A
B
A
B
A
B
A
B
P
A
P
B
A
B
A B
A 2
1
5
.
)
100
($
2
)
,
( 

 







Write out the lagrangian….we have three
constraints, so we have three multipliers
.5
1
2
.5 0
2 0
A
B
A P
P
 
 

  
  
Take derivatives with respect to ‘A’ and
‘B’ and set equal to zero
2 2
0 0 0
B B
 
  
1 1
0 0 0
A A
 
  
 
0 100 0 100 = 0
A B A B
P A P B P A P B
 
     
Possible Multiplier Conditions

67. 2 2
0 0 0
B B
 
  
1 1
0 0 0
A A
 
  
 
0 100 0 100 = 0
A B A B
P A P B P A P B
 
     
Possible Multiplier Conditions
We can simplify this down a bit....we already know that the constraint on income will bind
0 100 0
A B
P A P B
    
Further, consider the change in utility with respect to a change in apple consumption. Zero consumption of apples
would make this infinite, so the zero apple constraint will never bind
 
.5
.5
.5 .5 0
U
A
A




    1
0 =0
A 


68. .5
2
2
.5 0
2 0
100
0
0
A
B
A B
A P
P
P A P B
B

 


 
  
 


Now, let’s solve for the conditions under which the zero
banana condition binds
100
A
A
P

.5
.5
A
A
P



2 2 0
B
P
 
  
.5
.5
2 0
B
A
A
P
P

 
 
 
 
40
B A
P P


69. B
P
A
P
A
B P
P 40

Constraint is Non-Binding
Constraint is Binding
A
P
A
B
100
$
0


B
A
P
A
P
B


100
$
2
25
. 








A
B
P
P
A
Once again, prices determine whether or not the constraint is binding!