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

Glad that we got the same answer lulu. But why is there a different answer here-- http://answers.yahoo.com/question/index?qid=20071005203101AAUbSVu

?? And that person got 8 votes for it. Weird..

Answer 1

v = (1/3)pi(h^3/9)
v = (1/27)pih^3
dv/dt = (dv/dh)(dh/dt) = q - 10,000 cm^3/min.
q - 10,000 cm^3/min. = [(1/9)πh^2](20 cm/min.)
q - 10,000 cm^3/min. = [(1/9)π200^2](20 cm/min.)
q = 800000π/9 + 10,000
q ≈ 289,253 cm^3/min. ≈ 289.253 L/min.

Answer 2

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
[with you til there - not sure what you are doing next tho'
At this point, I would use dh/dt = 0.2 at h = 2; then you have K(pumped in rate) - 0.01m^3 = answer you find - hence find K]
dv/dt = [pi* 40000 * 20] / 9
dv/dt=279252 cm^3/min