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Math Help - Quotient Rule Word Problem

  1. #1
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    Post Quotient Rule Word Problem

    I am having difficulty remembering how to multiply the square root portion of this problem.

    A study indicates that spending money on pollution control is effective up to a point but eventually becomes wasteful. Suppose it is known that when x million dollars is spent on controlling pollution, the percentage of pollution removed is given by:

    P(x) = 100(square root x)/0.03x^2+9

    Determine the expenditure that results in the largest percentage of pollution removal. What is this maximum percentage?

    According to the quotient rule, I have (0.3x^2+9)(100square root x)-(100square root x)(0.06x)/(0.03x^2+9)^2

    Multipling out (3x+900square root x)-(6x^3/2)/0.03x^2+9)

    If I am right so far, can you give me a nudge on what to do with all these square roots?

    Thank You
    Last edited by becky; August 1st 2006 at 05:41 PM. Reason: incorrect problem
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  2. #2
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    Quote Originally Posted by becky

    P(x) = 100(square root x)/0.03x^2+9

    Determine the expenditure that results in the largest percentage of pollution removal. What is this maximum percentage?
    To avoid using constants I will do,
    P=\frac{Ax^{1/2}}{Bx^2+C}
    Where,
    A=100,B=.03,C=9

    To find the maximimumumumumum you need the critical points.
    Begin by finding the derivative,
    P'=\frac{(Ax^{1/2})'(Bx^2+C)-(Ax^{1/2})(Bx^2+C)'}{(Bx^2+C)^2}
    Thus,
    P'=\frac{\frac{1}{2}Ax^{-1/2}(Bx^2+C)-(Ax^{1/2})(2Bx)}{(Bx^2+C)^2}
    Since, the derivative exists everywhere we need when the derivative is zero, that is,
    \frac{1}{2}Ax^{-1/2}(Bx^2+C)-(Ax^{1/2})(2Bx)=0
    Since x^{-1/2}\not = 0 divide both sides by it,
    \frac{1}{2}Ax^{-1/2-(-1/2)}(Bx^2+C)-Ax^{1/2-(-1/2)}(2Bx)=0
    Simplify,
    \frac{1}{2}Ax^0(Bx^2+C)-(Ax^1)(2Bx)=0
    Thus,
    \frac{1}{2}A(Bx^2+C)-2ABx^2=0
    Open sesame,
    \frac{1}{2}ABx^2+\frac{1}{2}AC-2ABx^2=0
    Multiply by 2 and divide by "A":
    Bx^2+C-4Bx^2=0
    Thus,
    Bx^2-4Bx^2=-C
    Swictch signs,
    4Bx^2-Bx^2=C
    Thus,
    3Bx^2=C
    Thus,
    x^2=\frac{C}{3B}
    Thus, (substitute values back inside)
    x=\pm \sqrt{\frac{C}{3B}}=\pm \sqrt{\frac{9}{3(.03)}}}=\pm \sqrt{\frac{3}{\frac{3}{100}}}=\pm \sqrt{100}=\pm 10
    Reject the negative thus,
    x=10.

    (Now the trick is to show that 10 is a maximum )
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  3. #3
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    Hello, becky!

    You differentiated incorrectly . . .


    A study indicates that spending money on pollution control is effective up to a point
    but eventually becomes wasteful.
    Suppose it is known that when x million dollars is spent on controlling pollution,
    the percentage of pollution removed is given by: . P(x) \:= \:\frac{100\sqrt{x}}{0.03x^2+9}

    Determine the expenditure that results in the largest percentage of pollution removal.
    What is this maximum percentage?

    We have: . P(x)\;=\;\frac{100x^{\frac{1}{2}}}{0.03x^2 + 9}

    Then: . P'(x)\;=\;\frac{(0.03x^2 + 9)\cdot100\cdot\frac{1}{2}x^{-\frac{1}{2}} - 100x^{\frac{1}{2}}(0.06x)}{(0.03x^2 + 9)^2}

    . . . . . P'(x) \;= \;\frac{100\cdot\frac{1}{2}x^{-\frac{1}{2}}(0.03x^2 + 9) - 6x^{\frac{3}{2}}}{(0.03x^2+9)^2}


    Multiply top and bottom by 2x^{\frac{1}{2}}:\;P'(x) \;= \;\frac{100(0.03x^2 + 9) - 12x^2}{2x^{\frac{1}{2}}(0.03x^2+9)^2}

    . . P'(x) \;= \;\frac{3x^2 + 900 - 12x^2}{2\sqrt{x}(0.03x^2+9)^2}\;=\;\frac{900 - 9x^2}{2\sqrt{x}(0.03x^2+9)^2}  \;= \;\frac{9(100 - x^2)}{2\sqrt{x}(0.03x^2+9)^2}


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