A local company manufactures parts for a washing machine. It's marketing department finds that if the company manufactures x parts per month, its costs are

C(x) = 300x + 2500, and it's revenue are R(x) = -2x2 + 500x

We know that Profit = Revenue - Cost (and we hope revenue is more than cost!)

Determine what number of parts should be produced to maximize profit and what profit they will earn at this level.

Round your answer to the nearest dollar, but don't use the $ symbol.

Number of Parts

Maximum Profit

Question

A local company manufactures parts for a washing machine. It's marketing department finds that if the company manufactures x parts per month, its costs are

 C(x) = 300x + 2500, and it's revenue are R(x) = -2x2 + 500x

We know that Profit = Revenue - Cost  (and we hope revenue is more than cost!)

Determine what number of parts should be produced to maximize profit  and  what profit they will earn at this level.

Round your answer to the nearest dollar, but don't use the $ symbol. 

 Number of Parts   

Maximum Profit

whpalmer4 · Answer

Ah, I like these problems :-)

Profit = Revenue - Cost

$$P(x) = R(x) - C(x) = (-2x^2+500x) - (300x+2500)$$$$P(x) = -2x^2+200x-2000$$
$P(x)$ is a parabola, opening downward, so the vertex will be the maximum.  As the equation is written in the form $y = ax^2+bx+c$ we can find the x coordinate of the vertex (and thus the number of parts to be produced for maximum profit) via the expression $x = -\frac{b}{2a}$.  Evaluate $P(-b/2a)$ to find the maximum profit.