Costs
1. Cost Minimization (Ch 20)
2. Cost Curves (Ch 21)
Short-run total costs
 The short-run cost-minimization problem is
min w 1x1  w 2x 2
x1  0
f ( x1 , x 2 )  y .
 Note x2 is fixed at x2 = x2’
2
Short run costs: an example
 Consider the Cobb Douglas case again:
y  f(x 1 , x 2 )  x x .
1/2 1/2
1
2
 Let x2 be fixed at 4.
 Substituting x2 = 4 above we get
y  x 4  2x
1/2
1
1/2
1/2
1
 Rearranging this gives the short run
demand function for x1
y2
x1 
4
3
Short run costs: an example
 The conditional demand function for x1 and
the short run cost function are given by (a)
and (b) respectively
2
y
x (w1 , w 2 , y) 
4
(a)
S
1
2
(b)
y
c ( w1 , w2 , y )  w1 ( )  4 w2
4
s
4
Computing different costs
 Let w1 = 4 and w2 = 1/4. Then
c ( w1 , w2 , y )  c ( y )  y  1
s
2
Fixed costs: c(0) = 1
Average fixed costs:AFC(y) = c(0)/y = 1/y
Variable costs: y 2
Average variable costs: AVC(y) =
Variable cost/y = y
 Average costs: AC(y) = AVC + AFC =
y + (1/y)
 Marginal costs: MC(y) = dc(y)/dy = 2y




5
Diagram: AVC, AC, MC for the example
6
Diagram: AFC, AVC, AC in general
7
AVC, AC, and MC
8
Relationship between AC and MC.
 Note in the diagrams
- When AC increase with q, MC > AC
- When AC decreases with q, MC <AC
 Intuition: Suppose AC for 10 cars is
$15,000. If AC for 11 cars is $15,250,
then additional cost for the 11th car
(i.e. MC in this case) must be greater
than $15,000. That is MC > 15,000
On the other hand if AC for 11 cars is
14, 200 then 11 th car must cost less
than $15,000. That is MC < 15,000
9
Algebra: MC & AC
c(y)
AC(y) 
,
y
dC(y)
y
 (1 c(y))
 AC(y)
dy

.
2
y
y
Therefore,


 AC(y)
0
y
as





 AC(y)
0
y
y  MC(y)  c(y).
as
c(y)
MC(y) 
 AC(y).
y

10
Cost minimization with multiple plants
 A car manufacturing firm has two plants
with cost function as follows
Plant 1
c(y1 )  y1  1
Plant 2
c(y 2 )  2y 2  1
2
2
 Marginal costs
Plant 1
Plant 2
MC ( y1 )  2 y1
MC ( y 2 )  4 y 2
11
Cost minimization with multiple plants
Suppose target output is 100 cars
 Question:How do you split production
between between two plants?
 Answer: In a way such that
MC in plant 1 = MC in plant 2 (Why?)
 Suppose y1 = 75, y2 = 25.
 MC in plant 1 is 2 X 75 = 150 and
 MC in plant 2 is 4 X 25 = 100.
 Shift some production from plant 1 to plant
2. Do the reverse if MC in plant 2 is greater
than MC in plant 1
12
Cost minimization with multiple plants
 Solving for y1 and y2
MC(y1 )  MC(y 2 )
2y1  4y 2
y1  2y 2
 y1 + y2 = 100
 Solution: y1 = 66.66.., y2 = 33.33.
13
Cost minimization with multiple plants
 General Problem: Output target Y,
 How to split the output among n plants?
 Optimum:
MC(y1 )  MC(y 2 )  .....  MC(y n )
y1  y 2  ....  y n  Y
14
Special case: identical plants
 If all n cost functions are same, then equalizing
marginal costs give y1 = y2 = ….. = Y/n
 If the common cost function (for each plant) is
2
c( y )  y  1
 If a firm produces Y, each plant produces Y/n and
the total cost is
nc ( y )  n (Y / n )  n  (Y / n )  n
2
 Total cost function:
 Marginal cost:
2
C (Y )  (Y / n )  n
2
MC(Y) = 2Y/n
15
MC: Plant level and Firm level
16
From short run to long run
 Consider the Cobb Douglas case again:
y  f ( x1 , x2 )  x x .
1/ 2 1/ 2
1
2
 Let x2 be fixed, w1 = 1, w2 = 9
 The short run demand function for x1 in
terms of y and x2 is
y2
x1 
x2
2
y
c s ( w1 , w2 , y )  ( )  9 x2
x2
17
Long run: Choosing x2
 Cost function (as a function of x2)
2
y
c ( w1 , w2 , y )  ( )  9 x2
x2
s
 Suppose you have two choices x2 = 1 or
x2 = 2? Which one would you choose ?
18
Binary choice: x2 = 1 or 2?
 Target Output y = 2. What level of
x2 would you choose?
(i) Cost if x2 = 1: (4/1) + 9*1 = 13
(ii) Cost if x2 = 1: (4/2) + 9*2 = 20
Ans: Choose x2 = 1.
19
Binary Choice Again
 Everything is the same
 Target Output y = 6. What level of
x2 would you choose?
(i) Cost if x2 = 1: (36/1) + 9*1 = 45
(ii) Cost if x2 = 2: (36/2) + 9*2 = 36
Ans: Choose x2 = 2.
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Idea
 High x2 increases fixed cost, lowers
marginal cost.
 Bearing high fixed cost is worthwhile
only if reduction in marginal cost is
high enough (which happens when y
is high)
 If you are producing a lot of output
spending in machines is worthwhile
21
Continous choice of x2
 Problem: Choose x2 to minimize
y2
 9x2
x2
 Set
2
dg
y
  2 9  0
dx2
x2
 Optimal choice x2 = y/3
22
$/output unit
ACs(y;x2
)
ACs(y;x2 )
ACs(y;x2  )
The long-run av. total cost
AC(y)
curve is the lower envelope
of the short-run av. total cost curves.
y
23