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Math Help - differenciation

  1. #1
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    differenciation

    an offshore oil well is 2 kilometers off the coast. the refinery is 4 kilometers down the coast. laying pipe in ocean is twice as expensive as layiing it on land.
    what path should the pipe folow for the minimum cost?

    i got 6-√3 km from the refinery .. help me whether its right or nt please!
    Last edited by yugashan; April 24th 2013 at 09:28 AM.
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: differenciation

    No, that's incorrect. Can you show us how you set the problem up?
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  3. #3
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    Re: differenciation

    oh, sorry.. i got 7.47..
    c= land path + ocean path
    d=√(2^2-x^2)
    L=4-x
    c(x)= 2√(2^2-x^2)+(4-x)
    differenciation= ((2^2-x^2)^-1/2)-1
    when dc/dx=0 , x=-1
    c(-1) =7.47
    is it right?
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: differenciation

    I would draw a diagram first:

    differenciation-pipeline.jpg

    O represents the under ocean distance and L represents the over land distance. Let k represent the cost per unit length of the over land pipeline.

    By Pythagoras we find:

    x^2+2^2=O^2\,\therefore\,O=\sqrt{x^2+4}

    We also know:

    x+L=4\,\therefore\,L=4-x

    and so the cost function is:

    C(x)=2kO+kL=k\left(2\sqrt{x^2+4}+(4-x) \right)

    This is the function you want to minimize. What do you find when you equate the derivative to zero?
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