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Math Help - Cost Word Problem

  1. #1
    Junior Member shepherdm1270's Avatar
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    Cost Word Problem

    A company wishes to run a utility cable from point A on the shore to an installation at point B on the island. The island is 6 miles from the shore. It costs $400 per mile to run the cable on land and $500 per mile underwater. Assume that the cable starts at A and runs along ths shoreline, then angles and runs underwater to the island. Find the point at which the line should begin to angle in order to yield the minimum total cost.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by shepherdm1270 View Post
    A company wishes to run a utility cable from point A on the shore to an installation at point B on the island. The island is 6 miles from the shore. It costs $400 per mile to run the cable on land and $500 per mile underwater. Assume that the cable starts at A and runs along ths shoreline, then angles and runs underwater to the island. Find the point at which the line should begin to angle in order to yield the minimum total cost.

    Missing data, what you have provided is insufficient to solve the problem.

    RonL
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  3. #3
    Junior Member shepherdm1270's Avatar
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    We also did one like this in class and I have the notes that I can show you, maybe it will help:

    There's a guy who wants to get from point A to point B. Straight along a path is 300m and he can travel at 160m/min. The rest is woods at which he can travel at 70m/min and perpendicular from the path to point B is 800m. What route would minimize the amount of time?

    D = R*T
    T = D/R

    1 Scenario:
    E = 300m at 160 mpm = 18 min
    N = 800m at 70 mpm = 11.42 min

    So the total time would be straight along the path for 18 minutes, then up through the woods for 11.42 minutes...

    So what would minimize the amount of time?

    Pythagorean theorum: A^2 + B^2 = C^2

    So: what distance on the path (x)^2 plus distance up through the woods (800^2) = d^2

    So: x^2+800^2=d^2

    So D = SQRT (x^2 + 800^2)

    ....Total time on path is 300 (length of path) - x @ 160 mpm
    Total Tme is: 300-x/160 + SQRT(x^2+800^2)/70

    T'(x) = ... etc
    set = 0 for critical points
    get crazy number
    get 389 meters however this is not in the domain of [0, 300]
    so minimum has to occur at an endpoint.
    t(0) = 13.9 min
    t(300) = 12.83...

    so his best bet is traveling entirely through the woods





    OI!!!!!
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  4. #4
    Junior Member shepherdm1270's Avatar
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    and also, there is a picture i can't depict on here, with a value that was missing from the word problem. Distance on the shoreline from C to A is 9, distance directly up from C to B is 6...
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  5. #5
    Junior Member shepherdm1270's Avatar
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    can anyone help? i need an answer for monday, and i've worked on it several times to no avail... :-/ :-/
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  6. #6
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    Quote Originally Posted by shepherdm1270 View Post
    can anyone help? i need an answer for monday, and i've worked on it several times to no avail... :-/ :-/
    Note for the future: You say "i've worked on it several times to no avail... " Hmmmmm ..... It would be most helpful (mainly to yourself) if you showed this work that you've done.


    Let the required point along the shoreline be P.

    Let dPC = x so that dAP = 9-x.

    From Pythagoras: x^2 + 6^2 = (dPB)^2.

    Cost ($) = 400 \, dAP + 500 \, dPB = 400 (9 - x) + 500 \sqrt{x^2 + 36}.

    Differentiate the cost function with respect to x (use the chain rule), put the derivative equal to zero and solve for x:

    0 = -400 + \frac{500x}{\sqrt{x^2 + 36}}


    \Rightarrow 400 \sqrt{x^2 + 36} = 500 x


    \Rightarrow 16 (x^2 + 36) = 25 x^2 ......

    It's left to you to finish solving for x and then to test the nature of the solutions. You want the solution that corresponds to a minimum turning point. Call it x = p. Then the value of x that minimises the cost will be one of the following: p, 0, 9.

    It's left for you to think about where 0 and 9 come from and why they should be considered.
    Last edited by mr fantastic; March 29th 2008 at 10:09 PM.
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  7. #7
    Junior Member shepherdm1270's Avatar
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    So I solved for x, but I'm still confused where to take it from there. I know that x = 2.6666 or 2 2/3...

    i figured out that the distance between A and B when taking a direct route diagonially ( the C^2 in the pythagorean theorum) is 10.81

    Thus:
    Using endpoint 0 ( 9 on land, 6 underwater ) you get 3600 + 3000 = $6600
    Using endpoint 9 ( all underwater) or 10.81 under ($500) you get $5408.33

    Therefore I know it is cheaper to go completely underwater than from A to C then from C to B...

    however I'm not sure about the middle ones... :-\
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  8. #8
    Junior Member shepherdm1270's Avatar
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    never mind i figured it out. yaaaaaaaaaaaaaaaaaaaaay
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  9. #9
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    Quote Originally Posted by shepherdm1270 View Post
    So I solved for x, but I'm still confused where to take it from there. I know that x = 2.6666 or 2 2/3...

    i figured out that the distance between A and B when taking a direct route diagonially ( the C^2 in the pythagorean theorum) is 10.81

    Thus:
    Using endpoint 0 ( 9 on land, 6 underwater ) you get 3600 + 3000 = $6600
    Using endpoint 9 ( all underwater) or 10.81 under ($500) you get $5408.33

    Therefore I know it is cheaper to go completely underwater than from A to C then from C to B...

    however I'm not sure about the middle ones... :-\
    I think you'll find that x = 8. This is the value of x that gives the minimum cost of $5400. Your values of cost at x = 9 and x = 0 are correct and clearly gives costs greater than that at x = 8.
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