Expert: Abe Mantell - 7/8/2008

Question
Hi, I am having a problem solving a related rates problem.  
The height of a rectangular box is 10 inches.  Its length increases at a rate of 2 in/sec and its width decreases at the rate of 4 in/sec.  When the length is 8inches and the width is 6 inches, the volume of the box is changing atthe rate of?  
do i want to use the relation V=lwh?
i set up:
dV/dt to be found
dw/dt = -4
dl/dt = 2
is there a dh/dt and how do i find that if i am already looking for dV/dt?
I'm lost...any help is appreciated!  Thankyou.

Answer
Hello Stephanie,

Since h is constant, dh/dt=0 and no need to treat h as a variable.
So, V=L*W*10 or V=10LW...now differentiate w.r.t. "t"...
==> dV/dt= 10(L*dW/dt + dL/dt*W), product rule since both L & W are
functions of time (t).  Now substitute L=8, W=6, dL/dt=2, dW/dt=-4
==> dV/dt=10[8*(-4) + 2*6]=-200
So the volume is decreasing at the rate of 200 in^3/min.

OK?

Abe