Answer:
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
