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  1. #1
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    please help.

    The price p and the demand x are related by the equation x^2 p^2x =10. Suppose that the demand is increasing at the rate of 20 units/week. At what rate is p changing with respect to time when the demand is 50 units?
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    Quote Originally Posted by bobby77
    The price p and the demand x are related by the equation x^2 p^2x =10. Suppose that the demand is increasing at the rate of 20 units/week. At what rate is p changing with respect to time when the demand is 50 units?
    You know that x^2-p^2x=10. So take the time derivative of both sides:
    2x\frac{dx}{dt}-2xp\frac{dp}{dt}-p^2\frac{dx}{dt}=0.

    -Dan
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    please check my work...

    The price p and the demand x are related by the equation .x^2 p^2x =10 Suppose that the demand is increasing at the rate of 20 units /week. At what rate is p changing with respect to time when the demand is 50 units?
    Solution: x^2 p^2x =10 ---->(1)


    differentiating wrt time we get

    2x dx/dt -2xp dp/dt -p^2dx/dt =0
    we have
    dx/dt=20
    x=50

    substitute x=50 in eq(1)
    50^2 -p^2*50=10
    p= 7.05

    now to find dp/dt
    2*50*20 -2*50*7 dp/dt - 7^2 *20
    2000 -700 dp/dt -980=0
    700dp/dt= 1020
    dp/dt= 1020/700

    dp/dt= 1.457

    --------------------------------
    please check my work...
    thankyou
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by bobby77
    The price p and the demand x are related by the equation .x^2 p^2x =10 Suppose that the demand is increasing at the rate of 20 units /week. At what rate is p changing with respect to time when the demand is 50 units?
    Solution: x^2 p^2x =10 ---->(1)


    differentiating wrt time we get

    2x dx/dt -2xp dp/dt -p^2dx/dt =0
    we have
    dx/dt=20
    x=50

    substitute x=50 in eq(1)
    50^2 -p^2*50=10
    p= 7.05

    now to find dp/dt
    2*50*20 -2*50*7 dp/dt - 7^2 *20
    2000 -700 dp/dt -980=0
    700dp/dt= 1020
    dp/dt= 1020/700

    dp/dt= 1.457

    --------------------------------
    please check my work...
    thankyou
    This is basically OK except that you round p to 7.05 when it is ~7.0569,
    then you round it down again to exactly 7. This loses accuracy unnecessarily
    and gives your answer of dp/dt=1.457, when it should be more like 1.423.

    RonL


    PS I have deleted the duplicate post of this question.
    Last edited by CaptainBlack; March 1st 2006 at 03:02 AM.
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