A water tank is in the shape of an inverted cone with depth 10 meters, and top radius 8 meters. Water is flowing into the tank at a rate of 0.1 cubic meters / min, but leaking out at a rate of 0.002h^20.002 h 2 cubic meters / min, where hh is the depth of water in the tank, in meters. Find the depth of water when the volume of water in the tank is neither increasing nor decreasing.

The value of leaking rate in the question is repeated. By searching on the web I could find the correct value wich is 0.002h^2 m^3 /min.

The depth of the water has to be equal to 7.07 m in order to have a stationary volume.

Explanation:

In order to have a stationary water level the flow of water that comes into the tank (0.1 m^3/min) must be equal to the flow of water that goes out of the tank (0.002*h^2 m^3/min), therefore:

0.002*h^2 = 0.1

h^2 = 0.1/0.002

h^2 = 50

h = sqrt(50) = 7.07 m

correct it is c.) b b