Econ 301 chapter 21, technology and cost
Inputs are classified as “fixed” or variable.”
The distinction involves time
cost of fixed inputs = F
cost of variable inputs = cv(y)
total cost = c(y) = cv(y) + F
average cost at output level y is AC(y) = c(y)/y = cv(y)/y + F/y = AVC(y) + AFC
MC(y) = d c(y)/ d y = c’(y)
Facts that hold for all cost curves:
1. Cv(y) =∫MC(y) dy...which is the area under the MC curve.
2. If AC(y) has a minimum point, MC = ATC at min for ATC.
Proof : At bottom of AC, dAC/dy = (1/y2)(y c’(y) - c) = (1/y)(MC - ATC) = 0, so MC =
ATC.
3. At y = 0, AVC(y) = cv(y)/y = MC(y)...no proof necessary. (Use L’Hopital’s rule to
evaluate limit of cv(y)/y as y -> 0.)
Examples:
a. cost curves as in principles of micro
b. cost curves when y = a L, where a is positive constant
c. Cobb-Douglas ... see next page
100
Suppose we have the following
Cobb-Douglas production
function:
y = 10L2/3K1/3. This gives rise to
the isoquants on the left.
K
80
60
40
20
0
0
200
400
600
800
1000
L
Y
Suppose in short run K = 27, then
K1/3 = 3, and y = 30L2/3...this is
the short run labor production
function, and it looks like....
3000
2500
2000
1500
1000
500
200
400
600
800
1000
L
If you solve y = 30L2/3 for L, you get L = .0061y1.5, and this looks like...
L
150
100
50
Y
200
400
600
800
1000
wL = w(.0061)y1.5 is the total variable cost, which Varian writes as cv(y). If w = 8, cv(y)
would look like this....
$
1500
cv(y)
1250
1000
750
500
250
200
400
600
800
1000
Y
Total cost, which Varian writes as
c(y), is variable cost plus fixed cost, so c(y) = cv(y) + F. If fixed is $300, then the total
cost and variable cost look like this....
c(y)
1750
1500
1250
cv(y)
1000
750
500
250
200
400
The marginal cost is
600
MC( y ) 
800
1000
dc( y )
 (1.5)(8 )(.0061) y .5 ...It is the slope of c(y)
dy
and it looks like ...
$
MC(y)
2
1.5
1
0.5
200
400
600
800
1000
Y
The average total cost AC ( y ) 
put
the
ATC
and
c ( y ) cv( y ) F

  AVC ( y )  AFC( y ) ...if you
y
y
y
the
MC
together,
they
look
like
$
5
MC
4
3
ATC
2
1
1000
2000
3000
4000
5000
Y
this....
It is not hard to show that MC goes through the minimum point on ATC.. You can
identify the minimum point on ATC by taking the derivative of ATC and setting it to
zero. (You need to use the quotient rule.)
dATC

dy
d
c( y )
1
1
y
 ( 2 )( y  c '  c )  ( )( MC  ATC)
dy
y
y
The above equals zero, or ATC hits its minimum, only if MC = ATC.
Returns to scale:
For Cobb-Douglas, y = AKαLβ
it depends on α + β
increasing returns means α + β > 1
constant returns α + β = 1
decreasing returns α + β < 1
LRATC