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Math Help - Optimization money problem

  1. #1
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    Optimization money problem

    A variety store can sell 500 yo-yos, which cost the store 35 cents each, for $1 each. For each penny the store lowers the price it can sell 20 more yo-yo's. For what price should it sell the yo-yos to make a maximum profit??
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  2. #2
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    Quote Originally Posted by Mathforum View Post
    A variety store can sell 500 yo-yos, which cost the store 35 cents each, for $1 each. For each penny the store lowers the price it can sell 20 more yo-yo's. For what price should it sell the yo-yos to make a maximum profit??
    let x = number of 1 cent price reductions

    profit = revenue - cost

    profit = (number sold)(price each) - (number sold)(cost each)

    P = (500+20x)(1 - .01x) - (500+20x)(.35)

    maximize P w/r to x
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  3. #3
    Super Member Deadstar's Avatar
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    Quote Originally Posted by Mathforum View Post
    A variety store can sell 500 yo-yos, which cost the store 35 cents each, for $1 each. For each penny the store lowers the price it can sell 20 more yo-yo's. For what price should it sell the yo-yos to make a maximum profit??
    We express it as a function...

    Profit = Amountx(Sale Price) - Amountx0.35 = Amount((Sale Price) - 0.35)

    So we need mathematical representations for Sale Price and amount.

    Lets just call Sale Price x.

    We know that for x = 1, Amount = 500
    For x = 0.99, Amount = 520 etc

    So we model the amount sold as...

    500 + 2000(1-x)

    Why this?

    Well we start of with 500 at x=1, for every penny less, we add 20 more. So for x=0.98 we have +40 yoyos
    x = 0.97 we have +60 yoyos
    etc

    Lets look at 1-x to see what happens...

    for x=0.98, 1-x = 0.02 we have +40 yoyos
    for x = 97 we have 1-x = 0.03 and get +60 yoyos

    It can now been seen that 1-x multiplied by 2000 = the extra you sell.

    The formula also works when you raise the price.

    x=1.01,

    We get 500 - (0.01)2000 = 500 - 20 = 480 as anticipated.

    So anyway, We now have amount = 500 + 2000(1-x)

    Lets put this back into our original formula which we shall call P(x) where x was the sale price.

    P(Sale Price) = Amount((Sale Price) - 0.35)

    => P(x) = Amount(x - 0.35)

    => P(x) = (500 + 2000(1-x))(x - 0.35) (now lets expand it out)

    = 500x - 175 + 2000x(1-x) - 750(1-x)

    => P(x) = 3200x-875-2000x^2

    Find the derivative of this, set it equal to 0. Should be your answer.
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  4. #4
    Super Member Deadstar's Avatar
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    Quote Originally Posted by skeeter View Post
    let x = number of 1 cent price reductions

    profit = revenue - cost

    profit = (number sold)(price each) - (number sold)(cost each)

    P = (500+20x)(1 - .01x) - (500+20x)(.35)

    maximize P w/r to x



    Lol that's a lot simpler. Ah well...
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