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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- Mar 1st 2006, 12:34 AMbobby77please 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?

- Mar 1st 2006, 12:38 AMtopsquarkQuote:

Originally Posted by**bobby77**

$\displaystyle 2x\frac{dx}{dt}-2xp\frac{dp}{dt}-p^2\frac{dx}{dt}=0$.

-Dan - Mar 1st 2006, 02:05 AMbobby77please 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

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please check my work...

thankyou - Mar 1st 2006, 02:48 AMCaptainBlackQuote:

Originally Posted by**bobby77**

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.