Wednesday, August 5, 2015

Profit Maximization

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We have,
Profit = Revenue - cost
=> π = pq - (r1x1 + r2x2), where p = selling price
Say x2 = x2o (fixed)
 πo = pq - (r1x1 + r2x2)
=> q = (πo/p  +  r2x2o/p) + r1/p
Here, (πo/p  +  r2x2o/p) is intercept and r1/p is the slope.
This equation gives isoprofit lines.
Now, π =  pq - (r1x1 + r2x2)
 = pf(x1, x2) - (r1x1 + r2x2) [as q = f(x1, x2)]
∴ δπ/δx1 = pf1 - r1 = 0 -----(1)
 δπ/δx2 = pf2 - r2 = 0 --------(2)
from (1) & (2), pf1 = r1, pf2 = r2
So, the first order condition for profit maximization is the value of marginal product (pf1 or pf2) = price of the input (r1 or r2).
Again, pf1 = r1
=> p = r1/f1 and similarly p = r2/f2
∴ f1/f2 = r1/r2.
It is also the first order condition.

2nd order condition:
pf11 < 0, pf22 < 0  -----(3)
(3) implies that profit must be decreasing w.r.t further applications of either x1 or x2.

Math problem for profit maximization:
Given q = x1x2 and r1 = 4, r2 = 5, p =10
What is the maximum profit?
Solution:
From ist order condition of profit maximization,
f1/f2 = r1/r2
As q = x1x2
f1 = δq/δx1 = x2
f2 = δq/δx2 = x1
∴  x2/x1 = r1/r2
=> x2 = (4/5)x1
Again from 1st order condition,
pf1 = r1
=> px2 = 4
=> x2 = 4/p = 4/10 = 2/5 .
∴ x1 = (2/5). (5/4) = 1/2
∴  Profit, π = ...   = -2

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