Ask Question
24 July, 03:00

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?

+3
Answers (1)
  1. 24 July, 03:09
    0
    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
Know the Answer?
Not Sure About the Answer?
Find an answer to your 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 ...” in 📘 Mathematics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers