Answer:
32π ≈ 100.53 ft³/min
Step-by-step explanation:
You want the rate water is draining from an inverted cone when the level is 6 ft and dropping at 8 ft/min. The cone has a radius of 4 ft and a height of 12 ft.
Rate of change
The rate of change of volume is the product of the rate of change of height and the surface area of the water at the time of interest.
The radius of the cone is proportional to the height, so the radius of the water surface will be half the full radius when the height is half the full height. The surface radius when the height is 6 ft is ...
(6 ft)/(12 ft) × (4 ft) = 2 ft
The area of the water surface is ...
A = πr²
A = π(2 ft)² = 4π ft²
Then the rate of draining is ...
dV/dt = A(dh/dt) = (4π ft²)(-8 ft/min) = -32π ft³/min ≈ 100.53 ft³/min
The water is draining from the tank at about 100.53 cubic feet per minute when the water level is 6 ft.
__
Additional comment
The radius is 1/3 the height, so we can write the volume in terms of height as ...
V = 1/3πr²h = πh³/27
Then the rate of change is ...
V' = (πh²/9)h' = π(6²/9)(-8) = -32 π . . . . ft³/min
