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Math Help - Differential Calculus Problem

  1. #1
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    Differential Calculus Problem

    This is a question out of a textbook and it would be greatly appreciated if you could point me in the right direction. Don't worry, I've done most of it - I just need a little guidance on what to do next.

    =================================

    An oil platform is located at point O, at sea, 10km from nearest point P on a stretch of straight coastline. The oil refinery is at point R, 16km along the coast from point P (see diagram). A company needs to lay a pipeline from the platform to the refinery (ie. from point O to R). This is to consist of straight line sections. Laying pipeline underwater is 4/3 times as expensive as laying it on land:



    (a) If the pipeline reaches the coast at point X (as shown in my diagram) a distance x km from P in direction of R, find an expression for the cost C (in dollars) of laying the pipeline in terms of x, given that laying the pipeline on land has a fixed cost of A dollars per kilometre. Here's what I've done so far (could you check if it's right?):

    Using Pythagoras, OX = \sqrt{100 + x^2}

    C(x) = \frac{4}{3}\sqrt{100 + x^2} + (16 - x)

    (b) Find the cost of the least expensive route for laying the required pipeline. Here's what I've done:

    C(x) = \frac{4}{3}\sqrt{100 + x^2} + (16 - x)

    I differentiated it (hopefully I did it correctly):

    C'(x) = \frac{4x}{3\sqrt{100 + x^2}} - 1

    And simply solved for x (I used my calculator to do this):

    x = 11.339

    Next, I simply plugged this into my diagram to get the required dimensions:

    16 - 11.339 = 4.661

    So the required distance from X to R is 4.661 km.

    \sqrt{100 + 11.339^2} = 15.119

    So the required distance from O to X is 15.119 km.

    \frac{4}{3}(15.119) + 4.661 = 24.82

    Now what? Is this the answer? Obviously, it looks rather wrong. Am I supposed to put the dollar, A, somewhere? Please help...

    Thanks.
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  2. #2
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    Plug the value of x in the cost function you found at a to find the minimal costs
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  3. #3
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    C'(x) = \frac{4x}{3\sqrt{100 + x^2}} - 1

    "And simply solved for x (I used my calculator to do this)"
    Why not solve it algebraically?

    \frac{4x}{3\sqrt{100 + x^2}} - 1=0
    4x=3\sqrt{100 + x^2}
    16x^2=9(100 + x^2)
    7x^2=900
    x^2=\frac{900}{7}
    x=\frac{30}{\sqrt{7}}, as x should be positive.

    So your answer is C(\frac{30}{\sqrt{7}})
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  4. #4
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    Yup, that's what I did and I got 24.819.

    So the cost is about $24.82? Seems a little awkward (and a bit too cheap).
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  5. #5
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    No, the answer is $24.819*A. Sorry I read over it, you need to have;
    C(x) = A(\frac{4}{3}\sqrt{100 + x^2} + (16 - x)), so that you get C'(x) = A(\frac{4x}{3\sqrt{100 + x^2}} - 1). The optimal value for x stays the same.
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