Question

A rectangular box has a square base with an edge length of x cm and a height of h cm. The volume of the box is given by V = x2h cm3. Find the rate at which the volume of the box is changing when the edge length of the base is 12 cm, the edge length of the base is decreasing at a rate of 2 cm/min, the height of the box is 6 cm, and the height is increasing at a rate of 1 cm/min.

Answer 1

any idea?
what did you do so far...

Answer 2

we want to find \(\huge \frac{d}{dt}(V)\)

Answer 3

we have the volume
\(\huge V=x^2h\)
here both x and h are changing
you need to do product rule

Answer 4

\(\huge \frac{d}{dt}\left(V\right)=2x\frac{d}{dt}\left(x\right)h+x^2\frac{d}{dt}\left(h\right)\)

Answer 5

you are given the info you need
for x
d/dt (x)
h
d/dt (h)

Answer 6

just plugging the values

Answer 7

you need to be careful about decreasing rate of change

Answer 8

Refer to the Mathematica attachment.