Question on Related Rates and Cones?

Question

I was trying to solve a question and couldn't find the answer. Can someone please tell me where I am going wrong?

Question:
Water is leaking out of an inverted conical tank at a rate of 10000cm^3/min at the same time water is being pumped into the tank at a constant rate. The tank has height of 6m and the diameter at the top is 4m. If the water level is rising at a rate of 20cm/min when the height of the water is 2m, find the rate at which water is being pumped into the tank.

My answer:
h/6=r/2
r=h/3

v=(1/3)pi*r^2*h
v=(1/3)pi(h^3/9)

dv/dt = (pi*h^2)/9 * dh/dt
dv/dt = [pi* 40000 * 20] / 9
dv/dt=279252 cm^3/min

Can someone tell me where I am going wrong? Also, please do it using the same method and not the tan-vertex one.

IU

Answer 1

Your math is correct to that point. Your conclusion should be:
Rate of H2O pumped in - Rate of leak = dv/dt
Rate of H2O pumped in = dv/dt + Rate of leak
= 279252 + 10000 = 289252cm^3/min