   SEARCH HOME Math Central Quandaries & Queries  Question from Mary, a student: A rectangular box is 10 inches high. It's length increases at a rate of 2 inches per second and it's width decreases at the rate of 4 inches per second. When the length is 8 inches and the width is 6 inches, what is the rate of change of the volume? Hi Mary.

I have a similar problem.

Problem: My box is 7 inches high. Its length decreases at a rate of 1 inch per second and its width increases at a rate of 2 inches per second. When the length is 10 inches and the width is 5 inches, what is the rate of change of the volume?

Solution:
Volume is the product of Height, Length and Width: V = HLW. Since the Height is always 7 inches, V = 7LW.

The rate of change of the volume is the derivative of the volume with respect to time. So that is dV/dt. Thus we need the derivative of the left side of the volume function. So we take the derivative of both sides (whatever you do to one side of the equation you must do to the other):

dV/dt = d/dt ( 7LW )

Now I simplify this using the product rule:

dV/dt = 7 d/dt ( LW )
dV/dt = 7 [ L dW/dt + W dL/dt ]

Now look back at the problem: I know everything on the left side! L = 10 inches, dW/dt = -2 in/sec, W = 5 inches, and dL/dt = 1 in/sec. So I can just substitute in the values:

dV/dt = 7 [ (10)(-2) + (5)(1) ]
dV/dt = 7(-15)
dV/dt = -105 in3/sec

That's how I solved my problem. Can you use the same method to solve yours?

Cheers,
Stephen La Rocque.     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.