The base of a pyramid-shaped tank is a square with sides of length 12 feet, and the vertex of the pyramid is 10 feet above the base. The tank is filled to a depth of 4 feet, and water is flowing into the tank at the rate of 2 cubic feet per minute. Find the rate of change of the depth of water in the tank. (Hint: The volume of a pyramid is given by V=1/3Bh, where B is the base area and h is the height of the pyramid.)

I keep getting 0.0496 ft/min, but the answer key says 0.039 ft/min. Could someone go over this question step by step please?

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  1. volume of water is volume of pyramid minus the volume of pyramid above the water

    v = 480 - 480 (1 - d/10)²
    ...= 480 - 480 (1 - d/5 + d^2/100)
    ...= 96 d - 4.8 d^2

    dv/dt = 96 dd/dt - 9.6 d dd/dt

    2 = 96 dd/dt - 9.6 * 4 * dd/dt

    2 = 96 dd/dt - 38.4 dd/dt

    closer than .0496, but still not .039

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