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Math Help - Optimization Problem6

  1. #1
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    Optimization Problem6

    A boat leaves a dock at 2:00PM and travels due south at a speed of 20km/h.
    Another boat has been heading due east at 15km/h and reached the same dock at 3PM.
    At what time were the two boats closest together.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by camherokid View Post
    A boat leaves a dock at 2:00PM and travels due south at a speed of 20km/h.
    Another boat has been heading due east at 15km/h and reached the same dock at 3PM.
    At what time were the two boats closest together.
    Take the position of the dock as the origin, and 2 pm as t=0.

    The position of the first boat at time t is (0, -20t), and the position
    of the second boat at time t is (-15+15t, 0).

    hence the squate distance between them at time t is:

    SD = 15^2 (-1+t)^2 + 20^2 t^2.

    The distance between the boats is a minimum when the square distance is a
    mimimum, that is when d(SD)/dt = 0.

    d(SD)/dt = 2 15^2 (-1+t) + 2 20^2 t

    Which is zero when t = 9/25 hours, when SD=144, and so the seperation is
    12km.

    At this point we could use the second derivative test to show this is a
    minimum but that is probably not necessary here, as it must be.

    RonL
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  3. #3
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    Hello, camherokid!

    A slightly different approach . . .


    A boat leaves a dock at 2:00PM and travels due south at a speed of 20km/h.
    Another boat has been heading due east at 15km/h and reached the same dock at 3PM.
    At what time were the two boats closest together?
    Code:
          P   15t   B   15-15t    C
          * - - - - * - - - - - - *
                      \           |
                        \         |
                          \       |
                          x \     |20t
                              \   |
                                \ |
                                  |
                                  * A

    Ship #1 starts at C at 2 PM and travels south at 20 km/hr.
    In the next t hours, it travels 20t km to point A

    Since Ship #2 reaches C at 3 PM at 15 km/hr,
    . . it was 15 km west of C at 2 PM at point P.
    In the next t hours, it travels 15t km to point B.
    . . Hence: . BC = 15 - 15t


    Let x = AB, the distance between the ships.

    . . Then: .x .= .(20t) + (15 - 15t) .= .625t - 450t + 225

    Differentiate: .2x(dx/dt) .= .1250 t - 450

    . . and we have: .dx/dt .= .625t - 225)/x .= .0

    . . Hence: .t .= .9/25 .= .0.36 hours .= .21 minutes, 36 seconds


    Therefore, the ships are closest at 2:21'36" PM


    Since x .= .625(0.36) - 450(0.36) + 225 .= .144
    . . their minimum distance is: .x = 12 km.

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