Water is leaking out of an inverted conical tank at a rate of 0.0109 {\rm m}^3{\rm /min}. At the same time water is being pumped into the tank at a constant rate. The tank has height 9 meters and the diameter at the top is 4.5 meters. If the water level is rising at a rate of 0.24 m/min when the height of the water is 4 meters, find the rate at which water is being pumped into the tank. Must be accurate to the fifth decimal place.

Question

I have worked on and every time i get the same answer .43295

Answer 1

at a time of t min, let the height of the water level be h m, and the radius of the water level be r m
By ratios:
r/h = 4.5/9 = 1/2
2r = h
or
r = h/2

Volume of cone = (1/3)π r^2 h
= (1/3)π (h/2)^2 h
= (1/12)π h^3
dV/dt = (1/4)π h^2 dh/dt

given: when h = 4 m, dh/dt = .24 m/min

dV/dt = (1/4)π(16)(.24) = 3.015929
but that includes the leakage, so the actual
rate of change of the volume is
30.0159 + .0109 = appr 3.0268 m^3/min

check my arithmetic

Answer 2

if you do not know calculus
area of surface = pi d^2/4

area of surface * change in height/min = volume/minute
pi d^2/4 * .24 m/min = pump rate - leak rate