You know that . So take the time derivative of both sides:Originally Posted bybobby77

.

-Dan

Results 1 to 4 of 4

- Mar 1st 2006, 12:34 AM #1

- Joined
- Sep 2005
- Posts
- 136

- Mar 1st 2006, 12:38 AM #2

- Mar 1st 2006, 02:05 AM #3

- Joined
- Sep 2005
- Posts
- 136

## 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

- Mar 1st 2006, 02:48 AM #4

- Joined
- Nov 2005
- From
- someplace
- Posts
- 14,972
- Thanks
- 5

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.