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Math Help - optimzation problem #11

  1. #1
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    optimzation problem #11

    A lighthouse, L, is located on a small island 4 km west of point A on a straight north-south coastlne. A power cable is to be laid from L to the nearest source of power at point B on the shoreline, 12 km north of point A. The cost of laying cable under water is $6000/km and the cost of laying cable along the shoreline is $2000/km. To minimize the cost, the power line will be built from L underwater to a point C on the shoreline and then alone the shoreline from C to B. Find the location of point C (to the nearest metre) on the shoreline where the power cable should enter the water.
    Last edited by mr fantastic; August 12th 2009 at 03:53 PM. Reason: Changed post title
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  2. #2
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    Quote Originally Posted by skeske1234 View Post
    A lighthouse, L, is located on a small island 4 km west of point A on a straight north-south coastlne. A power cable is to be laid from L to the nearest source of power at point B on the shoreline, 12 km north of point A. The cost of laying cable under water is $6000/km and the cost of laying cable along the shoreline is $2000/km. To minimize the cost, the power line will be built from L underwater to a point C on the shoreline and then alone the shoreline from C to B. Find the location of point C (to the nearest metre) on the shoreline where the power cable should enter the water.
    Draw a diagram. Let C be x km north of A. Now give the cost of the cable c(x) in terms of x.

    Having got to this point you are left with a standard minimum problem; solve it.

    Ask for more help when you get stuch explaining exactly where you are having difficulties.

    CB
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    Draw a diagram. Let C be x km north of A. Now give the cost of the cable c(x) in terms of x.

    Having got to this point you are left with a standard minimum problem; solve it.

    Ask for more help when you get stuch explaining exactly where you are having difficulties.

    CB

    O.K.
    so far, I am at the part to find the domain of x.
    At this point, I believe that the domain of x is...
    0<x<12
    However, I am not 100% sure on this..
    When I did proceed with this domain I found that the min cost was 48,000 dollars per km when the cable is 12 km on the shoreline and 4 km underwater
    I dont know if this is correct.. can someone verify with me?
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  4. #4
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    At this point, I believe that the domain of x is
    Yes on this account.

    Drawing a diagram IS very helpful, especially when they tell you how they want the diagram to be drawn. If we have the LH to the west of A, and a point C between A and B (they tell us the best way to minimize our costs are if we run a cable underwater to the shore between A and B), we have a triangle:

    We can call the hypotenuse: |LC| which is the distance from the lighthouse to the point C between A and B
    We can call the base: |LA| defined as the shortest from the lighthouse to the shore, which is 4KM
    We can call the side opposite the Lighthouse: |AC|, which we can label X, which is the distance from A to C, and is the POINT on the shore where our underwater cable ends

    Above the Triangle LAC we have a single north-south line:

    We can call the distance from B to the point C: |BC| which is the point B to C, which is the cabling for the ground - we know the maximum this length can be is 12 (not really since we are told to get a point between A and B), and that any other point would be 12-X

    The next step would be to determine just what the relationships are between these sides/lines. From there, we can then form a function that gives us the cost of the underwater cable (the hypotenuse of our triangle, |LC|), and the shoreline-cable (|BC|).
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  5. #5
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    Quote Originally Posted by ANDS! View Post
    Yes on this account.

    Drawing a diagram IS very helpful, especially when they tell you how they want the diagram to be drawn. If we have the LH to the west of A, and a point C between A and B (they tell us the best way to minimize our costs are if we run a cable underwater to the shore between A and B), we have a triangle:

    We can call the hypotenuse: |LC| which is the distance from the lighthouse to the point C between A and B
    We can call the base: |LA| defined as the shortest from the lighthouse to the shore, which is 4KM
    We can call the side opposite the Lighthouse: |AC|, which we can label X, which is the distance from A to C, and is the POINT on the shore where our underwater cable ends

    Above the Triangle LAC we have a single north-south line:

    We can call the distance from B to the point C: |BC| which is the point B to C, which is the cabling for the ground - we know the maximum this length can be is 12 (not really since we are told to get a point between A and B), and that any other point would be 12-X

    The next step would be to determine just what the relationships are between these sides/lines. From there, we can then form a function that gives us the cost of the underwater cable (the hypotenuse of our triangle, |LC|), and the shoreline-cable (|BC|).
    So, is my answer above correct? min cost 48000 dollars when cable on shoreline is 12 km and cable underwater is 4 km?
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  6. #6
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    No. Using your measurements, that would mean they lay the cable underwater in a straight line from the Lighthouse to the point A, and then lay shoreline cable from A to B. Draw that on a piece of paper; you will see you get a right triangle. What is smaller the Hypotenuse, or the 12KM+4KM?

    $48,000 is going to be your maximum cost, but we want the OPTIMIZATION of this situation; in that we want the MINIMUM cost, and the corresponding distance of C from A that relates to that.

    The distance I got from A to C was \sqrt{2}. Hopefully I did that right, and perhaps someone can hop in and check.
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  7. #7
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    Quote Originally Posted by ANDS! View Post
    No. Using your measurements, that would mean they lay the cable underwater in a straight line from the Lighthouse to the point A, and then lay shoreline cable from A to B. Draw that on a piece of paper; you will see you get a right triangle. What is smaller the Hypotenuse, or the 12KM+4KM?

    $48,000 is going to be your maximum cost, but we want the OPTIMIZATION of this situation; in that we want the MINIMUM cost, and the corresponding distance of C from A that relates to that.

    The distance I got from A to C was \sqrt{2}. Hopefully I did that right, and perhaps someone can hop in and check.
    ok.. im still not too sure on this, or rather how you got it.
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  8. #8
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    This is actually difficult to explain w/o using a digram. Have you diagramed this out at least? That really needs to be your first step or none of this is going to make any sense if you can't visualize it.
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  9. #9
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    Quote Originally Posted by ANDS! View Post
    This is actually difficult to explain w/o using a digram. Have you diagramed this out at least? That really needs to be your first step or none of this is going to make any sense if you can't visualize it.
    yes.. i actually already did that when i was explaining to you my answer in the posts above. ^.^ what i really don't get though is how you obtain the min cost function for the variables to plug into the total cost function.
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  10. #10
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    Quote Originally Posted by skeske1234 View Post
    O.K.
    so far, I am at the part to find the domain of x.
    At this point, I believe that the domain of x is...
    0<x<12
    However, I am not 100% sure on this..
    No x may take any real value, but we do know that for the minimum cost it will be in the range 0 to 12 km.

    The diagram should look like the attachment, the heavy line represents the cable.

    optimzation problem #11-gash.png

    So the length of cable underwater is \sqrt{x^2+16} km and the length of cable on land is |12-x| km (but as we know the minimum cost will be achieved when x is in [0,12] we may drop the absolute value and look for the minimum in [0,12] )

    CB
    Last edited by CaptainBlack; August 11th 2009 at 07:59 PM.
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  11. #11
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    total cost:
    c(x)=(4+x)(6000)+(12-x)(2000)
    =4000x+48000
    c'(x)=4000

    what is wrong?
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  12. #12
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    Quote Originally Posted by skeske1234 View Post
    total cost:
    c(x)=(4+x)(6000)+(12-x)(2000)
    =4000x+48000
    c'(x)=4000

    what is wrong?
    \sqrt{x^2+16} \ne x+4
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