At a sand and gravel plant, sand is falling off a conveyor and onto a conical pile at a rate of 8 cubic feet per minute. The diameter of the base of the cone is approximately three times the altitude. At what rate (in ft/min) is the height of the pile changing when the pile is 12 feet high? (Hint: The formula for the volume of a cone is V = 1 3 πr2h.)

dh/dt = 0,008 ft/min

Step-by-step explanation:

The volume of the cone is:

V (c) = (1/3) * π*r²*h (1)

Where r is radius of the base

We know from problem statement

dV/dt = 8 ft³/min

And d = 3*h ⇒ 2*r = 3*h ⇒ r = (3/2) * h

Plugging the value of r in equation (1) we get

V = (1/3) * π*[ (3/2) * h ]²*h

V = (3/4) * π*h³

Now we differentiate relation to time, on both sides of the equation to get

dV/dt = (3/4) * π*3*h²*dh/dt

dV/dt = (9/4) * π*h²*dh/dt

The question is dh/dt when h = 12 ft. Therefore

8 = (9/4) * 3,14 * (12) ² * dh/dt ⇒ dh/dt = 8 / 1017,36

dh/dt = 0,008 ft/min