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Thread: 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 $\displaystyle \pi y$.
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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 $\displaystyle x$-by-$\displaystyle y$ rectangle with semicircles on the ends.



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

    The diameter of the semicircle is $\displaystyle y.$

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



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

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

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




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

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

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




    (d) Solve for $\displaystyle y.$

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


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




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

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

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

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




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

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

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

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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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