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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.

pf11 < 0, pf22 < 0 -----(3)

(3) implies that profit must be decreasing w.r.t further applications of either x1 or x2.

Given q = x1x2 and r1 = 4, r2 = 5, p =10

What is the maximum profit?

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

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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