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Math Help - Radius/Distane of a semicircular room...

  1. #1
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    Radius/Distane of a semicircular room...

    I need to determine the radius of the semicircular ends of the room in terms of Y. The perimeter of the room is to be a 200-meter single-lane running track.



    As well as finding the radius, I need to find the distance (in terms of Y) of the 2 semicircular parts of the track.

    Once those things are done, I have to take the radius and write an equation (in terms of x & y) for the distance traveled in one lap around the track.

    THEN I have to solve for Y. I then take that answer and write the area A of the rectangular region as a function of X.

    And FINALLY, I have to figure out t he dimensions that will produce a maximum area of the rectangle (x & y).

    I have absolutely no idea how to do any of this. Can't find this crap anywhere in the damn book.

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  2. #2
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    The distance around the whole circle is \pi y.
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  3. #3
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    Hello, BeSweeet!

    Come on!
    I'm sure you already know everything necessary to solve this problem.


    We have a 200-meter running track.
    The center is an x-by- y rectangle with semicircles on the ends.



    (a) Find the radius of the semicircular ends of the track in terms of y.

    The diameter of the semicircle is y.

    . . The radius is half of that: . \text{radius} \:=\:\frac{y}{2}



    (b) Find the distance (in terms of y) of the 2 semicircular parts of the track.
    The two semicircles make up one circle.

    Circumference of a circle: . C \:=\:2\pi r

    . . Therefore: . C \:=\:2\pi\left(\tfrac{y}{2}\right) \:=\:\pi y




    (c) Write an equation (in terms of x & y) for the distance around the track.

    \text{Distance} \;=\;\underbrace{\text{2 semicircles}}_{\pi y} + \underbrace{\text{2 straight tracks}}_{2x}

    . . d \;=\;\pi y + 2x




    (d) Solve for y.

    Since d = 200, we have: . \pi y + 2x \:=\:200 \quad\Rightarrow\quad \pi y \;=\;200 - 2x


    . . Therefore: . y \;=\;\frac{200-2x}{\pi} .[1]




    (e) Write the area A of the rectangular region as a function of x.

    The area of a rectangle is: . \text{Area} \:=\:\text{(length)} \times \text{(width)}

    So we have: . A \;=\;xy

    Substitute [1]: . A \:=\:x\left(\frac{200-2x}{\pi}\right) \quad\Rightarrow\quad A \:=\:\frac{1}{\pi}(200x - 2x^2)




    (f) Find the dimensions that will produce a maximum area of the rectangle.
    Solve A' \,=\,0

    A' \;=\;\frac{1}{\pi}(200-4x) \:=\:0 \quad\Rightarrow\quad 200 - 4x \:=\:0 \quad\Rightarrow\quad 4x \:=\:200

    . . Therefore: . x \:=\:50,\;y \:=\:\frac{100}{\pi}

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  4. #4
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    Wow, I get it now, except that some of those (I guess) were wrong:


    This is actually the last problem on this assignment. Can't wait to get it done!
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