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Thread: Quadratic equation

  1. #1
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    NWT, Canada
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    Question Quadratic equation

    When bicycles are sold for $300 each, a cycle store can sell 70 in a season. For every $25 increase in the price, the number sold drops by 10.
    a) Represent the sales revenue as a function of the price.
    b)The total sales revenue is $17,500. How many bicycles were sold? What is the price of one bicycle?
    c) What range of prices will give a sales revenue that exceeds $18,000?
    I'm stuck on b
    R= revenue
    n= number of bikes
    p= price
    For a
    Price| Number of Bikes sold
    300 | 70
    325 | 60
    350 | 50
    So find the slope
    $\displaystyle \frac{(70-50)}{(300-350)}=-0.4$
    Plug that into y=mx+b
    $\displaystyle 70=-0.4(300)+b$
    $\displaystyle 70=-120+b$
    $\displaystyle 70+120=190$
    $\displaystyle b=190$
    So that means the function to find n(number of bicycles sold) is $\displaystyle -0.4p+190$
    Since Revenue is $\displaystyle r=n*p$ then part a is $\displaystyle r=p(-0.4p+190)$ which simplifies to $\displaystyle r=-0.4p^2+190p$
    So to solve part b?

    $\displaystyle 17500=-0.4p^2+190p$
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  2. #2
    Junior Member
    Joined
    Feb 2008
    From
    NWT, Canada
    Posts
    51
    I got it
    $\displaystyle 17500=-0.4p^2+190p$
    $\displaystyle \frac{17500}{-0.4}=-43750$
    $\displaystyle -43750=p^2+(\frac{190}{-0.4}p$
    $\displaystyle 0=p^2-475p+43750$
    $\displaystyle 0=(p-125)(p-350)$

    $\displaystyle p=125, 350$
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