Question

The side length of a rectangular box with a square base is increasing at the rate of 2 ft/sec, while the height is decreasing at the rate of 2 ft/sec. At what rate is the volume of the box changing when the side length is 10 ft and the height is 12 ft?

680 ft3/sec
40 ft3/sec
-280 ft3/sec
280 ft3/sec

Answer 1

@freckles

Answer 2

If the base is fixed to be square as the side length increases, then the volume of the box is given by
\[V=x^2z\]
where \(x\) is the side length of the base and \(z\) is the height. Differentiating, you have
\[\frac{dV}{dt}=2x\frac{dx}{dt}z+x^2\frac{dz}{dt}\]
You're given that \(\dfrac{dx}{dt}=+2\) while \(\dfrac{dz}{dt}=-2\), and you want to find \(\dfrac{dV}{dt}\) at the time when \(x=10\) and \(z=12\).

Answer 3

actually I meant to close this question because I figured it out, but if you can help with my other question that I posted today I would be super thankful!

Answer 4

@SithsAndGiggles

Answer 5

@SithsAndGiggles