• March 1st 2006, 12:34 AM
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?
• March 1st 2006, 12:38 AM
topsquark
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
• March 1st 2006, 02:05 AM
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

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thankyou
• March 1st 2006, 02:48 AM
CaptainBlack
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

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