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Math Help - least distance

  1. #1
    Junior Member ihmth's Avatar
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    least distance

    1.) A ship lies 6 mi. from shore, and opposite a point 10 mi. farther along the shore another ship lies 18 mi. offshore. A boat from the first ship is to land a passenger and then proceed to the other ship. What is the least distance the boat can travel?

    Help pls...can anyone show me the diagram of this problem.


    -tnx
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  2. #2
    Super Member

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    Hello, ihmth!

    There is a back-door approach to this problem,
    . . involving optics and striaght-line paths.
    But I'll use the expected Calculus set-up.


    1) A ship lies 6 mi. from shore.
    Opposite a point 10 mi. farther along the shore another ship lies 18 mi. offshore.
    A boat from the first ship is to land a passenger and then proceed to the other ship.
    What is the least distance the boat can travel?
    Code:
                                          * C
                                        * |
                                      *   |
                                    *     |
        A *                       *       |
          | *                   *         | 16
          |   *               *           |
        6 |     *           *             |
          |       *       *               |
          |         *   *                 |
        B * - - - - - * - - - - - - - - - * D
                x     P       10 - x

    The first ship is A\!:\;AB = 6

    The second ship is at C\!:\;CD = 16
    And: . BD  = 10

    Let P be the landing point.
    Then: . BP \:=\: x\:\text{ and }\:PD \:=\:10-x

    In right triangle ABP\!:\;\;AP \:=\:\sqrt{x^2 + 6^2}

    In right triangle CDP\!:\;\;CP \:=\:\sqrt{(10-x)^2 + 18^2}

    The distance travelled is: . D \;=\;\left(x^2+36\right)^{\frac{1}{2}} + \left(x^2-20x+424\right)^{\frac{1}{2}}

    . . and that is the function we must minimize.

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