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Thread: quick help...please

  1. #1
    Oct 2007

    quick help...please

    A smokestack deposits soot on the ground with concentration inversely proportional to the square of the distance from the stack. With two smokestacks 20 miles apart, the concentration of the combined deposits on the line joining them, at a distance x from one stack, is given by the following equation where k1 and k2 are positive constants which depend on the quantity of smoke each stack is emitting.

    If k1 = 27k2, find the point on the line joining the stacks where the concentration of the deposit is a minimum.


    can someone please help me out with this one...much thanks

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  2. #2
    MHF Contributor
    Apr 2005
    S = [k1 /(x^2)] +[k2 / (20 -x)^2] ----(i)

    Since k1 = 27k2,

    S = [27k2 /(x^2)] +[k2 / (20 -x)^2] ----(ii)

    To find the minimum S, we equate dS/dx to zero.
    Remember, k1 and k2 are constants.

    Differentiate both sides of (ii) with respect to x,
    dS/dx = (27k2)[-2 /(x^3)] +(k2)[(-2 / (20 -x)^3) * (-1)]
    dS/dx = [-54k2 /(x^3)] +[2k2 / (20 -x)^3]

    Set that to zero,
    0 = [-54k2 /(x^3)] +[2k2 / (20 -x)^3]
    0 = (-54k2)(20 -x)^3 +(2k2)(x^3)
    0 = -54(20 -x)^3 +2x^3
    0 = -27(20 -x)^3 +x^3
    27(20 -x)^3 = x^3
    Get the cuberoots of both sides,
    3(20 -x) = x
    60 -3x = x
    60 = x +3x
    x = 60/4 = 15 miles
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