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Math Help - minimum value

  1. #1
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    minimum value

    need help...
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  2. #2
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    Hello, bobby,

    I label the island I.

    The total distance is d = IC + CD

    The total energy is e = 2*IC + CD

    Let BC = x

    Then you can calculate the energy by:

    e(x) = 2 \cdot \sqrt{5^2+x^2} + (13-x)

    You'll get the minimum if e'(x) = 0. So you need first the derivative of e (use chain rule!):

    e'(x) = 2 \cdot \frac{1}{2}\cdot \left( 5^2+x^2 \right)^{-\frac{1}{2}} \cdot 2x -1

    Now e'(x) = 0:

    2 \cdot \frac{1}{2}\cdot \left( 5^2+x^2 \right)^{-\frac{1}{2}} \cdot 2x -1=0
     \frac{2x}{\sqrt{25+x^2}} =1. Multiply both sides by the denominator and afterwards square both sides:
    4x^2 = 25+x^2. Solve for x. You'll get 2 values for x. The negative one is senseless in the described situation.

    I got x=\frac{5}{3}\cdot \sqrt{3}\approx 2.887 \text{ km}

    EB
    Last edited by earboth; November 6th 2006 at 12:57 AM.
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