Question

An electronics store sells 60 entertainment systems per month at $800 each. For every $20 increase in the selling price, the store sells one fewer system. What should the store set the price to in order to maximize the revenue? What is the maximum revenue? Provide an algebraic solution for full marks.

Answer 1

Let the number of price increases by x. Then since revenue = price * quantity,
r(x) = (800+20x)(60-x)
The vertex of that parabola is at (10,50000)
So max revenue of $50K when the price is 800+20*10 = $1000