An athletic store sells about 200 pairs of basketball shoes per month when it charges $120 per pair. For each $2 increase in price, the store sells two fewer pa
Mathematics
unicornspaulding
Question
An athletic store sells about 200 pairs of basketball shoes per month when it charges $120 per pair. For each $2 increase in price, the store sells two fewer pairs of shoes. How much should the store charge to maximize monthly revenue? What is the maximum monthly revenue?
1 Answer

1. User Answers Cecely
Let x = dollar increase in price
Let y = fewer number of pairs sold
Since 2 fewer shoes are sold for each 1 dollar (factor of 2)
y = 2x
Revenue = Number of shoes sold * Price charged per shoe
Number of shoes sold = 200  y = 200  2x
Price charged per shoe = $60 + $x
Revenue = (200  2x)(60 + x) = 2x^2 + 200x  120x + 12000
Revenue = 2x^2 + 80x + 12000
In a quadratic equation, Revenue is maximized when x = b / 2a. In this case:
x  80 / (2*2) = $20
Price charged per show = $60 + $x = $60 + $20 = $80.
Maximum revenue = 2x^2 + 80x + 12000 (evaluated at x = $20)
Maximum revenue = 2(20^2) + 80*20 + 12000 = $12800