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Math Help - Modeling Problem

  1. #16
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    Square about side because you getting ride of the root on the other correct?
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  2. #17
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    Quote Originally Posted by sk8erboyla2004 View Post
    Square about side because you getting ride of the root on the other correct?
    Yep - that's right.
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  3. #18
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    Starting off at my mistake


     (2l-l )^2= -130+l

     4l^2-80l-400=-130+l

    4l^2-80l-l=270

    l(4l-81)=270

    4l=351

    l = 87.75

    now what would I put for

    what path should he talk to minmize his cost and what was the min cost

    $87.75?
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  4. #19
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    Quote Originally Posted by danny arrigo View Post
    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).
    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
    Sorry, when I said it (**) was correct it isn't (should be 4000 instead of 400). Let me provide some details

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

    then

    C' = 10 - \frac{20(100-l)}{\sqrt{(100-l)^2+900}}

    Setting this equal to zero gives

     <br />
2(100-l) = \sqrt{(100-l)^2+900}\;\; \Rightarrow\;\;4(100-l)^2 = (100-l)^2 + 900<br />
so 100-l = \pm \frac{30}{\sqrt{3}}

    Thus l = 100 \pm 10\sqrt{3} where we choose the negative case as the postive gives to much.
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  5. #20
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    Quote Originally Posted by danny arrigo View Post
    Sorry, when I said it (**) was correct it isn't (should be 4000 instead of 400). Let me provide some details

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

    then

    C' = 10 - \frac{20(100-l)}{\sqrt{(100-l)^2+900}}

    Setting this equal to zero gives

     <br />
2(100-l) = \sqrt{(100-l)^2+900}\;\; \Rightarrow\;\;4(100-l)^2 = (100-l)^2 + 900<br />
so 100-l = \pm \frac{30}{\sqrt{3}}

    Thus l = 100 \pm 10\sqrt{3} where we choose the negative case as the postive gives to much.
    So l is the min cost and what should the path he should take be
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  6. #21
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    Quote Originally Posted by sk8erboyla2004 View Post
    So l is the min cost and what should the path he should take be
    Lay the pipeline l = 100-10\sqrt{3} = 82.679 ft. along the shoreline then cross into the river. The cost is C(100-10\sqrt{3}) = \$1,519.62. The endpoints ( l = 0, 100 actually give more).
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  7. #22
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    Just plug C

    100-10\sqrt{3}

    into the original cost function?
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  8. #23
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    Quote Originally Posted by sk8erboyla2004 View Post
    Just plug C

    100-10\sqrt{3}

    into the original cost function?
    Yep. That's what I did.
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  9. #24
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    what made you plug in the distance formula for 20l why not 10l or it would not have mattered?
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  10. #25
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    Quote Originally Posted by sk8erboyla2004 View Post
    what made you plug in the distance formula for 20l why not 10l or it would not have mattered?
    I'm sorry, I'm not sure of your question? Can you elaborate?
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  11. #26
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    Quote Originally Posted by danny arrigo View Post
    I'm sorry, I'm not sure of your question? Can you elaborate?
    I assume to cost function was

    10l+20l

    and you plugged in the distance formula for 20 l
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  12. #27
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    Quote Originally Posted by sk8erboyla2004 View Post
    I assume to cost function was

    10l+20l

    and you plugged in the distance formula for 20 l
    The cost function is C = 10 l + 20\sqrt{(100-l)^2 + 30^2} established earlier.
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  13. #28
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    Quote Originally Posted by danny arrigo View Post
    The cost function is C = 10 l + 20\sqrt{(100-l)^2 + 30^2} established earlier.
    Yes I know but could you have done

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