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14 August, 00:14

At a sand and gravel plant, sand is falling off a conveyor and onto a conical pile at a rate of 8 cubic feet per minute. The diameter of the base of the cone is approximately three times the altitude. At what rate (in ft/min) is the height of the pile changing when the pile is 12 feet high? (Hint: The formula for the volume of a cone is V = 1 3 πr2h.)

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  1. 14 August, 02:02
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    dh/dt = 0,008 ft/min

    Step-by-step explanation:

    The volume of the cone is:

    V (c) = (1/3) * π*r²*h (1)

    Where r is radius of the base

    We know from problem statement

    dV/dt = 8 ft³/min

    And d = 3*h ⇒ 2*r = 3*h ⇒ r = (3/2) * h

    Plugging the value of r in equation (1) we get

    V = (1/3) * π*[ (3/2) * h ]²*h

    V = (3/4) * π*h³

    Now we differentiate relation to time, on both sides of the equation to get

    dV/dt = (3/4) * π*3*h²*dh/dt

    dV/dt = (9/4) * π*h²*dh/dt

    The question is dh/dt when h = 12 ft. Therefore

    8 = (9/4) * 3,14 * (12) ² * dh/dt ⇒ dh/dt = 8 / 1017,36

    dh/dt = 0,008 ft/min
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