Question

An electronics store sells an average of 60 entertainment systems per month at an average of $800 more than the cost price. For every $20 increase in the selling price, the store sells one fewer system. What amount over the cost price will maximize revenue?

Answer 1

kidda

Answer 2

The previous answer is a good answer and is correct, however it may not be readily evident as to how it works. So I thought it might be nice to demonstrate it with real numbers.

Since x = the number we take away from 60 it is easy to see that each time we raise x by one, we are adding 20 dollars to the sale price and removing one unit from total monthly sales. The Revenue then is simply the total units sold times the increased price.

Let's start with x = 8 and end with x = 12 and see what the revenue looks like:
====
P = (800 + 20 * x)
Q = (60 - x)
P * Q = Revenue
(60 - x) * (800 + (20 * x)) = Revenue
====
(60 - 8) * (800 + (20 * 8)) = 49,920
(60 - 9) * (800 + (20 * 9)) = 49,980
(60 - 10) * (800 + (20 * 10)) = 50,000
(60 - 11) * (800 + (20 * 11)) = 49,980
(60 - 12) * (800 + (20 * 12)) = 49,920

It is easy to see that it peaks where x = 10 and P = 1000

Answer 3

thanks :)