Related Rates Calculus Problem

[thistleBranch]

You and some friends are trapped inside a rectangular room that is leaking water from the ceiling filling up the room with water. Picking up a small cubic gift box (6 inches in length) you time the box fills in exactly 3 minutes. While you were timing the filling of the box, your friends measured the room to be 10 feet x 12 feet x 8 feet (LxWxH). Assuming no water is seeping out of the room, determine how fast the water level in the room is rising the moment the water in the room is 5 feet deep.

This is a homework problem I have been stuck on for quite some time. I attached a picture of what I could work out. I would appreciate some help. Thank you.

 

[MarkFL]

Ignoring the water displaced by you and your friends, I would express the volume of water as the cuboid:

[MATH]V=w\ell h[/MATH]
And the width \(w\) and length \(\ell\) will be constant, we may differentiate with respect to time \(t\) to obtain:

[MATH]\frac{dV}{dt}=w\ell\frac{dh}{dt}[/MATH]
which implies:

[MATH]\frac{dh}{dt}=\frac{1}{w\ell}\frac{dV}{dt}[/MATH]
Using the information provided regarding how fast the gift box was filled, we know:

[MATH]\frac{dV}{dt}=\frac{6^3}{3}\frac{\text{in}^3}{\text{min}}[/MATH]
So, what do we find regarding the time rate of change of the depth of the water?

 

[thistleBranch]

Does this mean that
[MATH]\frac{dh}{dt} = \frac{1}{wl}(\frac{6^3}{3})\frac{in^3}{min}[/MATH]

 

[MarkFL]

Yes, plug in for the given width and length, then simplify to get some number in inches per minute.

 

[thistleBranch]

Thank you so much!