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Math Help - Maximizing Revenue!

  1. #1
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    Exclamation Maximizing Revenue!

    If the price charged for a candy bar is p(x) cents, then x thousand candy bars will be sold in a certain city, where p(x) = 39 - . How many candy bars must be sold to maximize revenue?

    I thought I could solve this problem by using implicit differentiation
    but don't know how to... Please help.
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  2. #2
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    Quote Originally Posted by Matho View Post
    where p(x) = 39 - .
    What is that supposed to mean? I dont get it, sorry
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  3. #3
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    Quote Originally Posted by Rapha View Post
    What is that supposed to mean? I dont get it, sorry
    That is 39 - x/24
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  4. #4
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    Quote Originally Posted by Matho View Post
    That is 39 - x/24
    Ok.

    p(x) = 39 - x/24 is the price for a candy bar
    you sell x candy bars, then your profit is

    x*p(x) = 39x - x^2 / 24 =: f(x)

    => f'(x) = 39 - 2*x^1 / 24 = 39 - x/12

    (f''(x) = - 1 /12 < 0 => Maximum )

    f'(x) = 0

    39 - x/12 = 0

    // + x/12

    39 = x/12

    // * 12

    x = 39 * 12 = 468

    f''(x) < 0 => x = 468 Maximum


    -----
    x*p(x) = 39x - x^2 / 24 =: f(x)
    ------

    f(468) = 468*(39 - 468 / 24) = 9126


    Im still confused about

    p(x) is the price charged for ONE candy bar, p(x) = 39 - x /24,

    but x thousand candy bars will be sold

    not sure about, maybe it is x*p(x)*1000, I dont know
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  5. #5
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    Thank you so much! That really helps me to solve this problem
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