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Math Help - Optimization Problems (time & distance)

  1. #1
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    Question Optimization Problems (time & distance)

    Hi everyone, I'm having some trouble with these 2 problems. Hope you can help me out.

    1. A toy tugboat is launched from the side of a pond and travels north at 5 cm/s. At the same moment, a toy launch starts from a point 8root2 m northeast of the tugboat and travels west at 7 cm/s. How closely to the two tugboats approach eachother.

    Answer: 186 cm

    2. Two isolated farms are situated 12 km apart on a straight country road that runs parallel to the main highway 20 km away. The power company decides to run a wire from the highway to a junctionbox, and from there, wires of equal length to the two houses. Where should the junction box be placed to minimize the length of wires needed?

    Answer: 16.5 km from the main road

    I can't seem to think of these questions the right way. Any help would be very appriciated.

    Thanks
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  2. #2
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    Quote Originally Posted by hsidhu View Post
    ...
    2. Two isolated farms are situated 12 km apart on a straight country road that runs parallel to the main highway 20 km away. The power company decides to run a wire from the highway to a junctionbox, and from there, wires of equal length to the two houses. Where should the junction box be placed to minimize the length of wires needed?

    Answer: 16.5 km from the main road

    ...
    1. Draw a rough sketch of the situation (see attachment)

    2. The total length of the used wire is:

    l = x + 2 w

    3. Since w^2 = 6^2+(20-x)^2 = 436-40x+x^2 you'l get:

    4. l(x)=x+2\cdot \sqrt{436-40x+x^2}

    5. Solve the equation l'(x) = 0 for x. (Don't forget to use the chainrule when deriving the function l)

    6. I've got x = 20 - 2\sqrt{3} \approx 16.5359
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  3. #3
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    Quote Originally Posted by hsidhu View Post
    Hi everyone, I'm having some trouble with these 2 problems. Hope you can help me out.

    1. A toy tugboat is launched from the side of a pond and travels north at 5 cm/s. At the same moment, a toy launch starts from a point 8root2 m northeast of the tugboat and travels west at 7 cm/s. How closely to the two tugboats approach eachother.

    Answer: 186 cm

    ...
    1. Draw a sketch (see attachment). The boats are sailing on the legs of an isosceles right triangle whose legs are 8 m long.

    2. Let t denote the elapsed time in seconds and transform the length of 8 m into 800 cm. The distance d is the hypotenuse of a right triangle with the legs (800 - 5t) and (800 - 7t) respectively.

    3. Thus:

    (d(t))^2=(800-5t)^2+(800-7t)^2 = 74t^2 - 19200t + 1280000

    4. If d(t) has an extreme value then the function D(t)=(d(t))^2 has a maximum. Consider the function D. Calculate the first derivation and solve D(t)' = 0 for t.

    5. Plug in this value into d(t)=\sqrt{74t^2 - 19200t + 1280000}.

    I've got d\left( \dfrac{4800}{37} \right) = \dfrac{800\sqrt{74}}{37} \approx 185.9962219
    Attached Thumbnails Attached Thumbnails Optimization Problems (time & distance)-tug_boats.png  
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  4. #4
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    Hello, hsidhu!

    The first one is tricky to set up . . .


    1. A toy tugboat is launched from the side of a pond and travels north at 5 cm/s.
    At the same moment, a toy launch starts from a point 8\sqrt{2} meters northeast
    of the tugboat and travels west at 7 cm/s.
    How closely do the two tugboats approach each other?

    Answer: 186 cm
    Code:
           800-7t L       7t
        B o - - - o - - - - - - - o C
          |      /              * |
          |     /             *   |
          |    / d          *     |
    800-5t|   /       _   *       |
          |  /    800√2 *         |
          | /         *           | 800
          |/        *             |
        T o       *               |
          |     *                 |
       5t |   *                   |
          | * 45                 |
        A o - - - - - - - - - - - o D
                    800

    The tugboat starts at A and goes north at 5 cm/sec.
    In t seconds, it travels 5t cm to T.

    The launch starts at C and goes west at 7 cm/sec.
    In t seconds, it travels 7t cm to L.

    Since AC = 800\sqrt{2}, it is the diagonal of an 800-cm square ABCD.
    . . AB = BC = CD = DA = 800

    Hence: . BT \:=\: 800-5t,\;\;BL \:=\: 800-7t

    We want to minimize distance: . d \:=\:TL


    In right triangle LBT\!:\;\;d \:=\:\sqrt{(800-5t)^2 + (800-7t)^2}

    Let D \:=\:d^2 \:=\:(800-5t)^2 + (800-7t)^2

    We have: . D \;=\;1,\!280,\!000 - 19,\!200t + 74t^2

    . . and that is the function we must minimize.

    Go for it!



    Edit: E.B. beat me to it . . . *sigh*
    .
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