Finding the Width of a Rectangle at a Changing Rate: A Related Rates Problem

[Swerting]

Homework Statement

A rectangle has a constant area of 200m² and its length L is increasing at the rate of 4 meters per second. Find the width W at the instant the width is decreasing at the rate of 0.5 meters per second.

Homework Equations

$$A=200$$
$$dA/dt =0$$ (since the area is constant)
$$dL/dt =4 m/s$$
$$dW/dt =-0.5 m/s$$
$$A=(L)(W)$$
and I'm not sure, but parameter is $$P=2L+2W$$

The Attempt at a Solution

I wrote L in terms of W, and W in terms of L, but I am having trouble taking L=(200/W) to dL/dt.
I know that most related rates problems need two equations, so I have been trying to figure out how parameters may work in. Any help is greatly appreciated.

 

[Pere Callahan]

$$dW/dt =-0.5 m/s$$

This is true at only one point in time, not as an identity. You first want to know the rate of change of W. Certainly, W=A/L, so
$$\frac{dW}{dt}=\frac{dW}{dL}\frac{dL}{dt}=-4\frac{A}{L^2}$$

For what value of L is this equal to .5? What is the corresponding value of W?

 

[Swerting]

Ahhhhh, I understand now!
My problem was just that, I thought that dW/dt=-.5 all the time, I forgot its connection to L! Thank you very much for your assistance.

 

[koosuns]

I'm sorry to bring this up again, but I have the same problem on a packet. I understand that dW/dt does not equal .5 all the time.

However, I don't see the reasoning behind "certainly, w = a/l, so..." and then the little graphic. Could someone explain it to me?