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Math Help - Optimization problem

  1. #1
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    Apr 2010
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    Optimization problem

    Hello, I am stuck on a problem and need a bit of help.

    An oil refinery is located on the north bank of a straight river that is 2 km wide. A pipeline is to be constructed from the refinery to storage tanks located on the south bank of the river 6 km east of the refinery. The cost of laying pipe is $400,000/km over land to a point P on the north bank and $800,000/km under the river to the tanks. To minimize the cost of the pipeline, where should the point P be located?

    Here is what I have so far.. (sorry for the drawing, I'm new to this and don't know of any better way to do it) EDIT: Drawing didn't transfer very well from the edit box


    I know that
    Cost = 400,000x + 800,000y

    and
    y = sqrt(2^2 +(6-x)^2)

    and I know that to find the min we need to find dC/dx and set equal to 0 but when I do this I get stuck when solving for x. Maybe I'm messing up my algebra, I don't know, but any help would be appreciated.

    Thanks
    Last edited by Zion; April 26th 2010 at 09:02 AM.
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  2. #2
    MHF Contributor
    Joined
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    Quote Originally Posted by Zion View Post
    Hello, I am stuck on a problem and need a bit of help.

    An oil refinery is located on the north bank of a straight river that is 2 km wide. A pipeline is to be constructed from the refinery to storage tanks located on the south bank of the river 6 km east of the refinery. The cost of laying pipe is $400,000/km over land to a point P on the north bank and $800,000/km under the river to the tanks. To minimize the cost of the pipeline, where should the point P be located?

    Here is what I have so far.. (sorry for the drawing, I'm new to this and don't know of any better way to do it) EDIT: Drawing didn't transfer very well from the edit box


    I know that
    Cost = 400,000x + 800,000y

    and
    y = sqrt(2^2 +(6-x)^2)

    and I know that to find the min we need to find dC/dx and set equal to 0 but when I do this I get stuck when solving for x. Maybe I'm messing up my algebra, I don't know, but any help would be appreciated.

    Thanks
    Hi Zion,

    for simplicity let the cost be 4x+8y hundred thousand dollars.

    Then, differentiating your equation,

    \frac{d}{dx}\left(8\sqrt{(6-x)^2+4}+4x\right)=8\frac{du}{dx}\  \frac{d}{du}u^{\frac{1}{2}}+4

    =4+(8)\frac{1}{2}u^{-\frac{1}{2}}\ \frac{dv}{dx}\ \frac{d}{dv}v^2=4+4\ \frac{1}{\sqrt{(6-x)^2+4}}(-1)(2)(6-x)

    where

    u=(6-x)^2+4

    v=6-x

    Setting this derivative to zero gives

    4+4\ \frac{1}{\sqrt{(6-x)^2+4}}2(x-6)=0

    \frac{8(x-6)}{\sqrt{(6-x)^2+4}}=-4

    \frac{8(6-x)}{\sqrt{(6-x)^2+4}}=4

    2(6-x)=\sqrt{(6-x)^2+4}

    4(6-x)^2=(6-x)^2+4

    3(6-x)^2=4

    (6-x)^2=\frac{4}{3}

    6-x=\frac{2}{\sqrt{3}}

    x=6-\frac{2}{\sqrt{3}}=\frac{6\sqrt{3}-2}{\sqrt{3}}

    Now you can calculate y from (6-x)^2=\frac{4}{3}

    and calculate the minimum cost from these x and y.
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  3. #3
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    Thanks so much!

    I was on the right track but kept getting tripped up in my algebra
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