Driving Short-run Cost Function from Cobb-Douglas Production
Function

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 = Æ’ (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

4Q^{2 }=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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