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Math Help - Demand Equation

  1. #1
    Newbie
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    Oct 2005
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    Demand Equation

    The price p and the quantity x sold of a certain product obey the demand equation

    p=-1/3x + 100 0 less then or equal x less then or equal 300

    a) Express the revenue R as a function of x.
    b) What is the revenue if 100 units are sold?
    c) What quantity x maximizes revenue? What is the maximum revenue?
    d) What price should the company charge to maximize.

    I am not sure how to get the answers. But I have the answers to this question. I am looking for someone to help me work it out. I have about 4 more like this to do. I am hoping someone can show me how to do this one and I can do the rest.

    Answers:
    a) R(x)=-1/3x^2 + 100x
    b) $6667
    c) 150; $7500
    d) $50

    robert
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  2. #2
    Junior Member
    Joined
    Oct 2005
    Posts
    36
    a) It looks like Revenue = Price * Quantity (Sound familiar to you? I have never seen this before....)
    So Revenue = px
    But p = -x/3 + 100
    So R(x) = -x/3 + 100x

    b) It is saying 100 units are sold. So x = 100. You just put this into the equation that you just worked out.

    R(100) = -100/3 + 100x100 = 6667

    c) To find out the maximum value the best way is to differentiate the equation and calculate dR/dx
    dR/dx = -2x/3 + 100
    Then set this equal to zero.
    -2x/3 + 100 = 0
    And you will find that x = 150

    For the second part just put 150 back into the the forumla for R(x).

    d) To see what price they need, just put the same value of x into the equation for p.

    Ok?
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