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? 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)
Q4Q2 =7,200
⸫ Q = (7,200/4)^1/2 = 42.43
Identical value MC and ATC obtained,
MC = 8 * 42.43 = 339.4
ATC= 4(42.43) + 7,200/42.43