Suppose that the profit (in hundreds of dollars) of a certain firm is approximated by: P(x, y) = 1500 +36x - 1.5x^2 + 120y - 2y^2
where x is the cost of a unit of labor and y is the cost of a unit of goods. Find the values of x and y that mazimize profit. Find the maximum profit.

Step-by-step explanation:.

Given the profit (in hundreds of dollars) of a certain firm is approximated by:

P(x,y) = 1500 +36x - 1.5x^2 + 120y - 2y^2 where

x is the cost of a unit of labor

y is the cost of a unit of goods.

The maximum value of x and y occurs when dP/dx = 0 and dP/dy = 0

dP/dx is the differential of the function with respect to x taking y as a constant

dP/dy is the differential of the function with respect to y taking x as a constant

dP/dx = 0+36-2(1.5)x^{2-1}+0-0

dP/dx = 36-3x

If dP/dx = 0

36-3x = 0

36-3x-36 = 0-36

-3x = -36

x = -36/-3

x = 12

Similarly;

dP/dy = 0+0-0+120-2(2)y^{2-1}

dP/dy = 120-4y

Since dP/dy = 0

120-4y = 0

Subtract 120 from both sides

120-4y-120 = 0-120

-4y = -120

y = -120/-4

y = 30

Hence the values of x and y that maximize the profit are 12 and 30 respectively.

To find the maximum profit, we will substitute x = 12 and y = 30 into the given function

P(x,y) = 1500 +36x - 1.5x^2 + 120y - 2y^2

P(12,60) = 1500 +36(12) - 1.5(12)^2 + 120(30) - 2(30)^2

P(12,60) = 1500 +36(12) - 1.5(144) + 120(30) - 2(900)

P(12,60) = 1500 + 432 - 216 + 3600 - 1800

P(12,60) = 1932-216+1800

P(12,60) = 3,516

Since the profit is in hundreds if dollars,

P(12,60) = 3516×100

P(12,60) = $351,600

Hence the maximum profit is $351,600

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