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Math Help - Crazily Difficult Optimization problem Please help

  1. #1
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    Crazily Difficult Optimization problem Please help

    An offshore oil well is located in the ocean at a point W, which is 5 miles from the closest point on the shore A on a straight line shoreline. the oil is to be piped to a point B in the shore that is 8 miles from A, by piping it on a straight line underwater to some point on the shore P between A and B and then on to B via pipe along the shoreline. if the cost of laying pipe is $100,000 per mile under water and $75,000 per mile over land, where should point P be located to minimize the cost of the laying pipe?

    Help will be very much appreciated, I have no idea where to start or what to do.
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    Quote Originally Posted by wair View Post
    An offshore oil well is located in the ocean at a point W, which is 5 miles from the closest point on the shore A on a straight line shoreline. the oil is to be piped to a point B in the shore that is 8 miles from A, by piping it on a straight line underwater to some point on the shore P between A and B and then on to B via pipe along the shoreline. if the cost of laying pipe is $100,000 per mile under water and $75,000 per mile over land, where should point P be located to minimize the cost of the laying pipe?

    Help will be very much appreciated, I have no idea where to start or what to do.
    how about starting with a sketch?

    let x = distance from point A to point P

    let C = cost in thousands of $ as a function of x

    C(x) = \sqrt{x^2 + 5^2} \cdot 100 +  (8-x) \cdot 75

    find the value of x that minimizes the cost
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    That helped a whole bunch. But how did you figure out C(x) ? Also the problem says that B is 8 miles from A why do you have an 8 - x from P to B ? After all this to find the minimum will I have to integrate C(x) ? If so how would I go about finding the values of my upper and lower limits ?
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    ohh never mind the 8 - x.
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    C(x) = (water distance)(cost of water pipe) + (land distance)(cost of land pipe)

    A to P = x

    P to B = 8-x
    -------------- add'em up
    A to B = 8

    you use derivatives to find maximums and minimums, not integrals.
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    Thank You very much.
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    Ok I sometimes have a problem graphing a function on my calculator so i was just wondering if the critical point is 0.15. If anyone might be able to confrim it that would be great.
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  8. #8
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    Quote Originally Posted by wair View Post
    Ok I sometimes have a problem graphing a function on my calculator so i was just wondering if the critical point is 0.15. If anyone might be able to confrim it that would be great.
    calculator? put that thing away until the very end.

    \displaystyle C'(x) = \frac{100x}{\sqrt{x^2+25}} - 75 = 0

    4x = 3\sqrt{x^2+25}

    16x^2 = 9(x^2+25)

    7x^2 - 225 = 0

    \displaystyle x = \frac{15}{\sqrt{7}} \approx 5.67
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    Thank you. When I fixed up my derivative i put the square root of x^2 + 25 on top. I neglected the sign of the power.
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