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Thread: [SOLVED] Price demand exquation

  1. Jun 26th 2009, 09:16 AM #1
    snappyslow22
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    [SOLVED] Price demand exquation

    I am having problems with the following problem that relates p(price) and demand (x) with the equation:

    <br /> x^2+2xp+25p^2=74,500<br />

    1. If the price is increasing at a rate of $ 2 per mo. when the price is $ 30; how do I find the rate of change of the demand?

    I understand that I need to implicitly dif. (2x*x'+2xp'+2px'+50pp') but I'm not sure what to plug in after this step...

    2. How do I find the rate of change of the price when demand is decreasing 6 units per mo when demand is 150?

    Any help would be greatly appreciated
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  2. Jun 26th 2009, 09:31 AM #2
    calc101
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    You plug in the following:

    p' = $2
    p = $30

    2. You implicitly differentiate, and then plug in:
    x' = 6
    and equate the equation to: 150

    I am not certain of my answers.
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  3. Jun 26th 2009, 09:36 AM #3
    snappyslow22
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    so for 1) I would end up with

    <br /> 2x*x'+2x(2)+2(30)+50(3)(2)=0<br />
    <br /> 2x*x'+4x+360=0<br />

    what do I do with the x'?
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  4. Jun 26th 2009, 09:49 AM #4
    borulce
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    you can plug in the first equation for p=$30 then you find x.
    with this you can solve the last equation you have found above.
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  5. Jun 26th 2009, 10:04 AM #5
    snappyslow22
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    I'm confused, can you demonstrate what you mean?
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  6. Jun 26th 2009, 10:07 AM #6
    Amer
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    Quote Originally Posted by snappyslow22 View Post
    I am having problems with the following problem that relates p(price) and demand (x) with the equation:

    <br /> x^2+2xp+25p^2=74,500<br />

    1. If the price is increasing at a rate of $ 2 per mo. when the price is $ 30; how do I find the rate of change of the demand?

    I understand that I need to implicitly dif. (2x*x'+2xp'+2px'+50pp') but I'm not sure what to plug in after this step...

    2. How do I find the rate of change of the price when demand is decreasing 6 units per mo when demand is 150?

    Any help would be greatly appreciated
    first find x when p=30

    x^2+60x+25(30)^2 = 74500

    x^2+60x+25(900)-74500 =0

    x^2+60x+100(225-745)=0

    x^2+60x+100(-520)=0

    x=\frac{-60\mp\sqrt{3600-(4(-52000))}}{2}

    x=\frac{-60\mp 10\sqrt{36+4(520)}}{2}

    x=\frac{-60\mp 10(2)\sqrt{529}}{2}

    x=\frac{-60\mp 20(23)}{2}

    x=\frac{20(-3+23)}{2}... or.....x=\frac{20(-3-23)}{2}

    you find the value of x when p=30 now sub it in the formula that you mention before you have the value of x and p and p' you can find x'
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  7. Jun 26th 2009, 10:17 AM #7
    borulce
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    x^2+2xp+25p^2=74,500
    for p=30 you can find x from the equation when the price is increasing at a rate of $ 2 per mo.

    last equation you found 2x*x'+4x+360=0 now you have x and you can find the rate of change of the demand namely x'
    I hope I could make it clear.
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