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**galactus** May I show you a short cut to these types of related rates problems?.

By making the observation that the rate of change of volume is equal to the cross-sectional area at that instant times the rate of change of the height.

$\displaystyle \frac{dV}{dt}=A(t)\cdot\frac{dh}{dt}$

$\displaystyle \frac{dh}{dt}=\frac{\frac{dV}{dt}}{A(t)}$

We know that dV/dt=4.

The area of the water surface, A(t), when it is 4 feet deep is just a rectangle with area 16*4=64.

So, we have $\displaystyle \frac{dh}{dt}=\frac{4}{64}=\frac{1}{16}$