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Math Help - Modeling Problem

  1. #1
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    Modeling Problem

    Ive encounter this problem which is very confusing for me

    Tim needs to run a pipeline to his camp that is 100 feet west along the shore and 30 feet north across a lake from his starting position. It cost $10 per foot to run pipe over land and $20 to run pipe over water What path should Tim take to minimize his cost? What his minimum cost ?


    I so if I draw a mental picture you a height of 30 feet on the y axis and 100 feet on the x axis

    I know I can use the distance formula somehow but i am still lost
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    Quote Originally Posted by sk8erboyla2004 View Post
    Ive encounter this problem which is very confusing for me

    Tim needs to run a pipeline to his camp that is 100 feet west along the shore and 30 feet north across a lake from his starting position. It cost $10 per foot to run pipe over land and $20 to run pipe over water What path should Tim take to minimize his cost? What his minimum cost ?


    I so if I draw a mental picture you a height of 30 feet on the y axis and 100 feet on the x axis

    I know I can use the distance formula somehow but i am still lost
    Suppose you move l ft in the x direction. The cost for this would be C = 10 l. Now the distance between you (now) and the point across the river is \sqrt{(100-l)^2 + 30^2} and the cost for this portion is C = 20\sqrt{(100-l)^2 + 30^2} and thus the total cost is

    C = 10 l + 20\sqrt{(100-l)^2 + 30^2}

    Now minimize C (i.e. \frac{dC}{dl} = 0 and find l). Don't forget to check the endpoints l = 0 (all river) and l = 100 (all shore then across).
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    For this did you use the distance formula and what else
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    Quote Originally Posted by sk8erboyla2004 View Post
    For this did you use the distance formula and what else
    For the cost function? Yes, the distance forumla and the cost per length.
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    so take the derivative and solve for 0
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    Quote Originally Posted by sk8erboyla2004 View Post
    so take the derivative and solve for 0
    Yep!
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    Ok whew I took the derivative which took a few good lines of work here is my final solution for the derv.

    If this is correct I will continue on and solve this for 0


    10+\frac{40l-400}{2\sqrt{(100-l)^2+900)}}

    ??? I am so unsure of this
    Last edited by sk8erboyla2004; May 2nd 2009 at 06:27 PM.
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    Quote Originally Posted by sk8erboyla2004 View Post
    Ive encounter this problem which is very confusing for me

    Tim needs to run a pipeline to his camp that is 100 feet west along the shore and 30 feet north across a lake from his starting position. It cost $10 per foot to run pipe over land and $20 to run pipe over water What path should Tim take to minimize his cost? What his minimum cost ?


    I so if I draw a mental picture you a height of 30 feet on the y axis and 100 feet on the x axis

    I know I can use the distance formula somehow but i am still lost
    There is a similar problem in this thread: http://www.mathhelpforum.com/math-he...s-problem.html

    Most of these sorts of optimisation problems have been asked and answered several times in the forums over the years. It's always worth doing a search of key words ....
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    Thanks however is my differentiation correct?
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    Quote Originally Posted by sk8erboyla2004 View Post
    Thanks however is my differentiation correct?
    Looks goods so far but cancel before solving for l.
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    Cancel what?
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  12. #12
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    Quote Originally Posted by sk8erboyla2004 View Post
    Ok whew I took the derivative which took a few good lines of work here is my final solution for the derv.

    If this is correct I will continue on and solve this for 0


    10+\frac{40l-400}{2\sqrt{(100-l)^2+900}}

    ??? I am so unsure of this
    From you post

    C' = 10+\frac{40l-400}{2\sqrt{(100-l)^2+900}} = 0

    so

     10 \left( 1+\frac{2l-20}{\sqrt{(100-l)^2+900}} \right) = 0

    so

     1+\frac{2l-20}{\sqrt{(100-l)^2+900}} = 0
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    damn thats gonna be hard to solve :?
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    Quote Originally Posted by danny arrigo View Post
    From you post

    C' = 10+\frac{40l-400}{2\sqrt{(100-l)^2+900}} = 0

    so

     10 \left( 1+\frac{2l-20}{\sqrt{(100-l)^2+900}} \right) = 0

    so

     1+\frac{2l-20}{\sqrt{(100-l)^2+900}} = 0

    Ok If I solve for l I do

     \frac{2l-20}{\sqrt{(100-l)^2+900}} = -1

     {2l-20} = -1(\sqrt{(100-l)^2+900}

     2l-20 = -100+l-30

     2l-l = -130+20

     l = -110

    ? ehhh
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  15. #15
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    Quote Originally Posted by sk8erboyla2004 View Post
    Ok If I solve for l I do

     \frac{2l-20}{\sqrt{(100-l)^2+900}} = -1

     {2l-20} = -1(\sqrt{(100-l)^2+900}

    \color{red}{2l-20 = -100+l-30}

     2l-l = -130+20

     l = -110

    ? ehhh
    The step in red isn't right. I'll need to square both sides.
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