Water is leaking out of an inverted conical tank at a rate of 6,500 cm3/min at the same time that water is being pumped into the tank at a constant rate. The tank has height 6 m and the diameter at the top is 4 m. If the water level is rising at a rate of 20 cm/min when the height of the water is 2 m, find the rate at which water is being pumped into the tank. (Round your answer to the nearest integer.) cm3/min

285,753 cm³/min

Step-by-step explanation:

The rate of change of volume is the product of the water's surface area and its rate of change of depth.

At a depth of 2 m, the water has filled 1/3 of the 6 m depth of the tank. So, the radius at the water's surface will be 1/3 of the tank's radius of 2 m. The water's surface area is ...

A = πr² = π (2/3 m) ² = 4π/9 m²

The rate of change of depth is 0.2 m/min, so the volume of water is increasing at the rate ...

dV/dt = (0.20 m/min) (4π/9 m²) = 8π/90 m³/min ≈ 279253 cm³/min

This change in volume is the difference between the rate at which water is being pumped in and the rate at which it is leaking out:

2.8*10⁵ cm³/min = (input rate) - 6500 cm³/min

Adding 6500 cm³/min to the equation, we get ...

input rate ≈ 285,753 cm³/min