Mathematics

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

  • 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


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