Calculus
posted by Martin on .
A leaky cylindrical oilcan has a diameter of 4 inches and a height of 6 inches. The can is full of oil and is leaking at the rate of 2 cubic inches/hr. The oil leaks into an empty conical cup with a diameter of 8 inches and a height of 8 inches.
a. At what rate is the depth of the oil in the conical cup rising when the oil in the cup is 3 inches deep?
b. When the oilcan is empty, what is the depth of the oil in the conical cup?
Please show stepbystep work in your answers. Thanks in advance! :)

let the height of oil in the cone be h in, let the radius of the surface of oil in the cone be r in. Let the volume in the cone be V in^3
by ratios: r/h = 4/8
r = h/2
given d(V)/d(time) = 2 cu inches/hour
find dh/dt when h = 3 inches
V = (1/3)pi(r^2)h subbing in r=h/2
V = (1/12)pi(h^3)
dV/dt = (1/4)pi(h^2)dh/dt
sub in dV/dt = 2 and h = 3 to get
dh/dt = 8/(9pi) inches/hour
for b)
find the volume of the cylinder.
sub that into V=(1/12)pi(h^3) and solve for h
Let me know if you got h = 6.6 
For part (b), I don't understand why h = 6.6 ft because that would mean that the volume of the oilcan had a volume of 75.27 in^3. In actuality, the oilcan had a volume of 25.13 in^3, so the height should be 4.6 ft. **

Ryan is wrong. Reiny is correct. h = 6.6

A drum can in the shape of a rectangular shape with a square base is required to have a volume of 100 cubic feet. Draw a diagram of the drum indicating x, the side of the base and h, the height.