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Math Help - REVENUE questions

  1. #1
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    REVENUE questions

    Denide is an artist who works at a shopping cneter drawing "pencil potraitts". SHe charges $20 per potraite and is averaging 30 potraits per week. she decides to increase the price , but every 1 doller increase she will lose one sale per week. it takes her $10 to make one potrait. wat should she set the price to get the max. profi.
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  2. #2
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    Quote Originally Posted by usm_67
    Denide is an artist who works at a shopping cneter drawing "pencil potraitts". SHe charges $20 per potraite and is averaging 30 potraits per week. she decides to increase the price , but every 1 doller increase she will lose one sale per week. it takes her $10 to make one potrait. wat should she set the price to get the max. profi.
    Let S(\Delta p) represent sales. Where \Delta p represents the change in price. For example, if \Delta p=1 means that the price was increased by 1 dollar, then S(\Delta p) represents the sales. Since as you said she loses a sale for each dollar you have that S(\Delta p)=29 because 30 without changing the price and 29 if increasing the price by 1 dollar.

    We can create the first few values for \Delta p and see how they affect the sales.
    \left\{ \begin{array}{cc} \Delta p&S(\Delta p)\\ 0&30\\1&29\\2&28\\3&37
    Do you see a formula? Yes,
    S(\Delta p)=30-\Delta p.

    Remember that,
    \mbox{Profit }=\mbox{ Revenue }-\mbox{ Cost}
    In this case if she has to pay $20 for each sale since there are 30-\Delta p sales in total we have that,
    \mbox{Cost }=20(30-\Delta p).
    Now, she recieves 20+\Delta p for every sale, because $20 was original and she increase the price by \Delta p
    But she makes 30-\Delta p sales thus in total she recieves (20+\Delta p)(30-\Delta p) thus,
    \mbox{Revenue }=(20+\Delta p)(30-\Delta p)
    Finally here profit is the difference between these values thus,
    (20+\Delta p)(30-\Delta p)-20(30-\Delta p)
    Open parantheses (use FOIL and watch those signs )
    30\Delta p-(\Delta p)^2
    This is a parabola with a maximum point the max occurs at, \Delta p=15 in that case she has to increase her price by 15 dollars thus the new price would be 35 dollars.
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