Answer:
x = 31 hundred dolars and
y = 91/2 = 45.5 hundred dolars
Step-by-step explanation:
Given
R(x) = (40−8x+5y)*x + (50+9x−7y)*y
C(x) = (40−8x+5y)*10 + (50+9x−7y)*29
We can use the equation
P(x) = R(x) - C(x)
where
P(x) is the profit
R(x) is the revenue
and C(x) is the costs
In order to maximize the telephone company's profit, we apply
P'(x) = R(x)' - C(x)' = 0
⇒ R(x)' = ((40−8x+5y)*x + (50+9x−7y)*y)' = (40x-8x²+14xy+50y-7y²)'
⇒ C(x)' = ((40−8x+5y)*10 + (50+9x−7y)*29)' = (1850+181x-153y)'
⇒ P'(x) = -8x²-7y²-141x+203y+14xy-1850
The first-order partial derivatives of these functions are
Px(x,y) = -16x-141+14y
Py(x,y) = -14y+203+14x
Setting these equal to zero and solving we obtain:
-16x+14y-141 = 0
14x-14y+203=0
we get the solution
x = 31 and y = 91/2 = 45.5
Finally, the company should produce 3100 units of the first system, and 4550 units of the second system.
