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19 January, 13:18

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.

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  1. 19 January, 16:40
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    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
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