I am getting something mixed up

Question

A trough is 12 feet long and 3 feet across the top (see figure). Its ends are isosceles triangles with altitudes of 3 feet. 

(a) If water is being pumped into the trough at 2 cubic feet per minute, how fast is the water level rising when h is 1.2 feet deep? 

If the water is rising at a rate of 3/8 inch per minute when h = 2.5, determine the rate at which water is being pumped into the trough. 

Answer 1

The volume water in the trough = 1/2 base*h*length where the base value is the width measured at the surface of the water. Since the base value is equal to the height of the water the formula for volume = 1/2h²L or V=6h² since L=12. Note that dV/dt=12hdh/dt using the chain rule and that dV/dt = 2 (given) and at h=1.5 dh/dt= 1/4 ft/min.