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26 February, 19:44

A box is partially filled with liquid. The length of the box is 11 inches. The width of the box is 17 inches. If the volume of the liquid is increasing at a rate of 562 cubic inches per minute, what is the rate, in inches per minute, at which the height of the liquid is changing when the height of the liquid is 4 inches?

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  1. 26 February, 22:45
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    dh/dt = 3.01 inches/minute

    The height of the liquid is changing at the rate of 3.01 inches/minute

    Step-by-step explanation:

    Volume of liquid in the box is;

    V = length * width * height

    V = lbh ... 1

    If the volume of the liquid is increasing at a rate of 562 cubic inches per minute.

    dV/dt = 562 cubic inches per minute

    Length l = 11 inches

    Width b = 17 inches

    From equation 1; differentiating the equation we have;

    dV/dt = bh (dl/dt) + lh (db/dt) + lb (dh/dt) ... 2

    Since the length and width of water is not changing then;

    dl/dt = db/dt = 0

    Equation 2 becomes;

    dV/dt = lb (dh/dt)

    dh/dt = (dV/dt) / lb

    Substituting the values;

    dh/dt = 562 / (11*17) inch/minute

    dh/dt = 3.01 inches/minute

    The height of the liquid is changing at the rate of 3.01 inches/minute
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