Water is leaking out of an inverted conical tank at a rate of $12000.0$ cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height $8.0$ meters and the diameter at the top is $3.5$ meters. If the water level is rising at a rate of $25.0$ centimeters per minute when the height of the water is $4.5$ meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute.
For this problem I am using "$R$" to represent the rate at which water is being pumped into the tank, and "$V$" to represent volume.
I know that $$\frac{dV}{dt} = R - 12000,$$ and that the volume of a cone is represented by $$\frac{1}{3}\pi r^2h.$$
How do you solve this?
Use the dimensions of the tank and similar triangles to replace $r$ with something in terms of $h$ in your equation $V=\frac13\pi r^2h$. So $V=f(h)$. Then $\frac{dV}{dt}=f'(h)\frac{dh}{dt}$ by the chain rule. And this should be enough help.