Water is leaking out of an inverted conical tank at a rate of 60000 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 90 meters and the diameter at the top is 40 meters. If the water level is rising at a rate of 300 centimeters per minute when the height of the water is 30 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute.

The key is writing volume as function of height of water.

Volumecone= 1/3 h*basearea=1/2*h(20/90)^2PI
take the derivative, dh/dt= 3m/min, find dV/dt

To solve this problem, we need to apply the principle of related rates. This involves finding equations that relate the variables in the problem and differentiating them with respect to time.

Let's denote the following variables:
- V: volume of water in the tank
- h: height of the water in the tank
- r: radius of the water surface in the tank

We are given the following information:
- The rate at which water is leaking out: dV/dt = -60,000 cm^3/min
- The rate at which the water level is rising: dh/dt = 300 cm/min

We need to find the rate at which water is being pumped into the tank: dV_pump/dt.

We can relate V, h, and r using the properties of a cone. The volume of a cone can be expressed as V = (1/3)πr^2h.

Since the diameter of the top of the tank is 40 meters, the radius is half of that: r = 20 meters.

Differentiating the equation of the volume with respect to time, we get:
dV/dt = (1/3)π(2rh)(dh/dt) + (1/3)πr^2(dh/dt)

We can now plug in the given values and solve for dV_pump/dt:

-60,000 = (1/3)π(2(20)(90))(300) + (1/3)π(20^2)(300)

Simplifying and solving for dV_pump/dt will give us the final answer.

Note: All dimensions used in the calculations should be in the same unit (e.g., meters or centimeters) to ensure consistency.