Driving Short-run Cost Function from Cobb-Douglas Production
Function
Normally, every firm face two
problems while producing their product,
1.
How
much quantity of goods to produce?
2.
And
how much labor and capital is required to produce output most efficiently?
The production Function state
relationship between output and input such as,
Q = ƒ (K, L)
Where,
Q is output or
quantity
K is the input of
capital
L is the input of
Labor
Cobb-Douglas
Production Function
Q = L^a K^b
Where,
Q is output
L is unit of
labor
K is a unit of
Capital
a & b = the parameter to be estimated.
Let a & b
equally = 0.5
Q = L^0.5
K^0.5
Suppose cost
components of a firm are as follow,
Price of a new
machine – K = $36 million
The wage of each
worker – W = $144 per day
Rental rate of
capital or price of capital – r = $200 per annum
“r” = (interest
rate + Rate of Depreciation) x Price of capital
Lets determine
the value of L
Q = L^0.5
36^0.5 = L^0.5 x 6
Therefore, L^0.5
= Q/6
L = (Q/6)^2
= Q^2/36
Total Cost (TC)
= Variable Cost (VC) + Fixed Cost (FC)
TC = W*L + r*K
TC = 144*Q^2/36 + 200*36
TC = 4Q^2
+ 7,200
VC = 4Q^2
FC = 7,200
Average total
cost (ATC) = TC / Q = 4Q^2/Q + 7,200/Q
Therefore, ATC
= 4Q + 7,200/Q
AVC = 4Q and
AFC = 7,200/Q
We can calculate
the Marginal Cost by taking first derivative of TC
TC = 4Q^2
+ 7,200
⸫ MC = dTC/dQ =
(4*2) Q^2-1 + 0 = 8Q
Minimum Cost condition
where,
MC = ATC
8Q = 4Q +
7,200/Q
8Q – 4Q =
7,200/Q
4Q = 7,200/Q
Multiply both
sides of equation by Q
Q(4Q) = (7,200/Q)
Q
4Q2 =7,200
⸫ Q = (7,200/4)^1/2
= 42.43
Identical value
MC and ATC obtained,
MC = 8 * 42.43
= 339.4
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