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Math Help - Optimation Problem

  1. #1
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    Optimation Problem

    A house (H) is located on an island that is 200 north of a point (A) on the shore. A Power station (p) is located on the shoreline at a point 400 m west of Point A. An electric cable is going to be laid from the power station to the house. It costs $12/m to lay underwater and $6/m to lay underground. How should the cable be placed so that the cost is a minimum and what is this minimum cost?

    Please help, exam is in 1.5 hour. Thanks
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  2. #2
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    What have you done so far?

    Clarification question: is that $12/m or $12/mm for the underwater cost?
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  3. #3
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    I cant seem to make a cost equation
    its $12/meter

    Thanks
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  4. #4
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    Well, to start you out, I'd go with

    \text{cost}=\$12d_{w}+\$6d_{g},

    where d_{w} is the length of underwater cable, and d_{g} the length of underground cable.

    So, I'm assuming you're going to want to go some distance east from the power station, underground, and then at some point you're going to go underwater. Is this correct?
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  5. #5
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    Thats looks correct and could you please elaborate on the equation, on how I could make that a derivative
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  6. #6
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    Ok. You need to find d_{g} and d_{w} in terms of one parameter: the point at which the cable leaves the shore. Draw a picture and start labeling things. What do you get?
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  7. #7
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    You description is pretty ambiguos. The interpretation i ahve used is in the attachment.

    Optimation Problem-diag5.jpg

    There are only 2 routes the cable is likely to take (p ->h), (p -> a -> h). Try them both and see which is cheaper.
    Last edited by mr fantastic; June 21st 2010 at 01:48 PM. Reason: No edit - Moved from thread created by duplicate post.
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  8. #8
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    Shouldn't it be that the cable goes a certain distance underground, and then underwater instead of going all the way to A first
    Last edited by mr fantastic; June 21st 2010 at 01:48 PM. Reason: No edit - Moved from thread created by duplicate post.
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  9. #9
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    Quote Originally Posted by SpringFan25 View Post
    You description is pretty ambiguos. The interpretation i ahve used is in the attachment.

    Click image for larger version. 

Name:	diag5.jpg 
Views:	11 
Size:	5.5 KB 
ID:	17949

    There are only 2 routes the cable is likely to take (p ->h), (p -> a -> h). Try them both and see which is cheaper.
    I like your disclaimer! And this is not accurate- there are an infinite number of possible routes and neither of the ones you give is likely to be optimal. Put the cable from H to some point X between H and A. Let "x" be the distance from H to X. The cost of laying that cable is 6x.

    The distance remaining from X to A is 200- x and the distance across from A to P is 400 so you can use the Pythagorean theorem to find the straight line distance form X to P. Multiply that by 12 to find the cost of laying that cable.

    The total cost is the sum of those two costs. Differentiate that sum with respect to x and set the derivative equal to 0 to find the x that gives optimal cost.

    By the way, the correct word is "optimization", not "optimation".
    Last edited by mr fantastic; June 21st 2010 at 01:49 PM. Reason: No edit - Moved from thread created by duplicate post.
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