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Math Help - [SOLVED] Optimization Problem argh

  1. #1
    falcarius
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    [SOLVED] Optimization Problem argh

    Hi everyone!

    The Question is:
    "A 400km racetrack is to be built with two straight sides and semicircles at the ends. Find the dimensions of the track that encloses the maximum area."
    Given:
    The two long sides of the rectangle are written with >/= to 100m (each)
    The straight side of the 2 semi circles is written with >/= to 20m (each)
    (>/= means greater than or equal to, just in case)

    Relevant equations are probably:
    Area of semicircle = 1/2 * pr2
    Area of rectangle = lw

    Well i tried but its always the beginning setting up of optimization problems that is the killer, the rest is always easy. It is a study Q for an upcoming test.
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  2. #2
    Newbie servantes135's Avatar
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    Quote Originally Posted by falcarius View Post
    Hi everyone!

    The Question is:
    "A 400km racetrack is to be built with two straight sides and semicircles at the ends. Find the dimensions of the track that encloses the maximum area."
    Given:
    The two long sides of the rectangle are written with >/= to 100m (each)
    The straight side of the 2 semi circles is written with >/= to 20m (each)
    (>/= means greater than or equal to, just in case)

    Relevant equations are probably:
    Area of semicircle = 1/2 * pr2
    Area of rectangle = lw

    Well i tried but its always the beginning setting up of optimization problems that is the killer, the rest is always easy. It is a study Q for an upcoming test.
    Falcarius, you are missing ^ with Aof S
    S=pi*(r)^2/2

    and in the end you will have 2 semi circles and 1 rectangle.
    Plus you also need to know
    Cs=pi*r

    the total circumprance will be 400km and
    Tc=2*(pi*r)*b where b is the base of the rectangle
    Tc=400km

    we want to maximize the total area
    Ta=(pi*r^2)+(2*r*b) since the hight of the rectangle is the same as the diamiter or twice the radius of the circle.

    with this

    400/pi=(2*r*b)*pi/pi
    400/pi=(2*r*b)

    substitute 400/pi to make
    f(r)=Ta=pi*r^2+400/pi

    from there it's a matter of finding what r is and then finding b
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  3. #3
    Newbie servantes135's Avatar
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    oh and a better way of making >/= symbol is to underline > or < like so < or > . I hope this helped
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  4. #4
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    Hello, falcarius!

    This is a trick question . . . certainly a tricky question.


    A 400-km racetrack is to be built with two straight sides and semicircles at the ends.
    Find the dimensions of the track that encloses the maximum area.

    Given: The two long sides of the rectangle > 100 m.
    The diameter of the 2 semicircles >20 m.

    I went through the expected procedure for an optimization.
    Code:
                  * * * * * * * * * * *
              *     |               |     *
            *       |               |       *
           *       r|               |r       *
                    |               |
          *         |               |         *
          *         +               +         *
          *         |               |         *
                    |               |
           *       r|               |r       *
            *       |               |       *
              *     |               |     *
                  * * * * * * * * * * *
                           L

    Let L = length of the straight tracks.
    Let r = radius of the semicircular tracks.

    The total length of the straight tracks is: 2L.
    The total length of the curved tracks is: 2\pi r
    . . Hence, we have: . 2L + 2\pi r \:=\:400\quad\Rightarrow\quad L \:=\:200 - \pi r . [1]


    The area of the rectangle is: 2rL.
    The area of the two semicircles is: \pi r^2.
    . . Hence, the total area is: . A \;=\;\pi r^2+ 2rL . [2]

    Substitute [1] into [2]: . A \;=\;\pi r^2+ 2r(200 - \pi r)\quad\Rightarrow\quad A \;=\;400r - \pi r^2

    Differentiate and equate to zero: . A' \;=\;400 - 2\pi r \;=\;0

    . . Hence: . r \,=\,\frac{200}{\pi}

    Substitute into [1] and we get: . L \,=\,0 .
    . . . What's going on?


    It is fairly well-known that, for a given perimeter,
    . . the figure with maximum area is a circle.
    So for maximum area, the track should be circular.

    But we are told that the straight tracks must be at least 100 meters (each).
    . . What do we do now?

    It can be shown that, as L increases, the area decreases.
    . . Hence, for maximum area, we will use the least L, 100.

    Under the restrictions, maximum area is attained when L = 100,\;r = \frac{100}{\pi}

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  5. #5
    Newbie servantes135's Avatar
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    dang I see my mistake. ty man
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