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Math Help - Optimization Problem

  1. #1
    Junior Member
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    Optimization Problem

    The director of a cruise ship company wants to transport passengers from the ship to a hotel 10 miles down the coast. To simplify the problem, assume the coast is straight. The perpendicular distance between the ship and the coast is 5 miles. The director plans to have power boats take the passengers to a point on the coast and then have minivans take them the rest of the way to the hotel. The power boats travel at a speed of 20 miles per hour while the minivans can go 35 miles per hour. To what point on the coast should the power boats take the passengers so that the travel time for the passengers is minimized. What is the minimum travel time?
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  2. #2
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by camherokid View Post
    The director of a cruise ship company wants to transport passengers from the ship to a hotel 10 miles down the coast. To simplify the problem, assume the coast is straight. The perpendicular distance between the ship and the coast is 5 miles. The director plans to have power boats take the passengers to a point on the coast and then have minivans take them the rest of the way to the hotel. The power boats travel at a speed of 20 miles per hour while the minivans can go 35 miles per hour. To what point on the coast should the power boats take the passengers so that the travel time for the passengers is minimized. What is the minimum travel time?
    Code:
          5 miles
       |------------|
     
      (x).          |            -
    (boat)  . 20mph |            |
               .    |            |
                  . |            |
            -       |(a)         | 10 miles
            |       | .          |
          d |       | . 35mph    |
            |       | .          |
            -       |(y) (hotel) -
    That was fun to draw .

    From the boat (x), the power boat will take the passengers to the shore at (a), then the minivan will take the passengers to the hotel (y).

    Let the distance between (a) and (y) be equal to d, then the distance from (x) to (a) is sqrt[5^2 + (10 - d)^2]

    The total time traveled will be:
    T = 20*sqrt[25 + (10 - d)^2] + 35*d

    Now we have a function relating distance and time. Taking the derivative, we get:
    T' = 20*1/2(25 + (10 - d)^2)^(-1/2)*(-2)(10 - d) + 35
    = -20*(10 - d)/sqrt[25 + (10 - d)^2] + 35

    Letting T' = 0:
    0 = -20*(10 - d)/sqrt[25 + (10 - d)^2] + 35
    35*sqrt[25 + (10 - d)^2] = 200 - 20d

    Sqare both sides:
    35^2*[25 + (10 - d)^2] = 40000 - 8000d + 400d^2
    1225(25 + 100 - 20d + d^2) = 40000 - 8000d + 400d^2
    153125 - 24500d + 1225d^2 = 40000 - 8000d + 400d^2
    825d^2 - 16500d + 113125 = 0
    25(33d^2 - 660d + 4525) = 0
    d = (660 +/- sqrt[435600 - 597300])/66

    I made an "oops" somewhere. There's too many digits! . Give me a few minutes to gain my compsure and maybe I'll figure this one out.
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  3. #3
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by ecMathGeek View Post
    Code:
          5 miles
       |------------|
     
      (x).          |            -
    (boat)  . 20mph |            |
               .    |            |
                  . |            |
            -       |(a)         | 10 miles
            |       | .          |
          d |       | . 35mph    |
            |       | .          |
            -       |(y) (hotel) -
    That was fun to draw .

    From the boat (x), the power boat will take the passengers to the shore at (a), then the minivan will take the passengers to the hotel (y).

    Let the distance between (a) and (y) be equal to d, then the distance from (x) to (a) is sqrt[5^2 + (10 - d)^2]

    The total time traveled will be:
    T = 20*sqrt[25 + (10 - d)^2] + 35*d

    Now we have a function relating distance and time. Taking the derivative, we get:
    T' = 20*1/2(25 + (10 - d)^2)^(-1/2)*(-2)(10 - d) + 35
    = -20*(10 - d)/sqrt[25 + (10 - d)^2] + 35

    Letting T' = 0:
    0 = -20*(10 - d)/sqrt[25 + (10 - d)^2] + 35
    35*sqrt[25 + (10 - d)^2] = 200 - 20d

    Sqare both sides:
    35^2*[25 + (10 - d)^2] = 40000 - 8000d + 400d^2
    1225(25 + 100 - 20d + d^2) = 40000 - 8000d + 400d^2
    153125 - 24500d + 1225d^2 = 40000 - 8000d + 400d^2
    825d^2 - 16500d + 113125 = 0
    25(33d^2 - 660d + 4525) = 0
    d = (660 +/- sqrt[435600 - 597300])/66

    I made an "oops" somewhere. There's too many digits! . Give me a few minutes to gain my compsure and maybe I'll figure this one out.
    I found my oops.

    T = D/R

    If you need me to go through the calculations (which are much easier now that I know what the equation should be), I can. But I think you can do the rest on your own.
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  4. #4
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    Lexington, MA (USA)
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    Hello, camherokid!

    The director of a cruise ship company wants to transport passengers
    from the ship to a hotel 10 miles down the coast.
    To simplify the problem, assume the coast is straight.
    The perpendicular distance between the ship and the coast is 5 miles.
    The director plans to have power boats take the passengers to a point on the coast
    and then have minivans take them the rest of the way to the hotel.
    The power boats travel at a speed of 20 mph while the minivans can go 35 mph.
    To what point on the coast should the power boats take the passengers
    so that the travel time for the passengers is minimized.
    What is the minimum travel time?
    Code:
          S
          *
          |  *
          |     *     _______
        5 |        * √x + 25
          |           *
          |              *
          * - - - - - - - - * - - - - - - -*
          A        x        P     10-x     H

    The ship is at S; the hotel is at H. .AP = 5, AH = 10.

    The power boat will land at P (x = AP) and drive to H (PH = 10 - x).

    . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ______
    Using Pythagorus, the distance on water is: .SP = √x + 25 miles.
    . . . . . . . . . . - . . . - . . . . . . . . . . . _______
    . . At 20 mph, this will take: .T
    1 .= .√x + 25 / 20 hours.


    The distance on land is: .PH .= .10 - x miles.
    . . At 35 mph, this will take: .T
    2 .= .(10 - x) / 35 hours.


    The total time is: .T . = . T
    1 + T2 . = . (1/20)(x + 25)^ + (2/7) - (1/35)x


    And that is the function we must minimize . . .

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