1. 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.

2. The distance around the whole circle is $\pi y$.

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

4. 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!

the figure given below shows a running track surrounding a grassed enclosure pqrstu. the enclosure consists of a rectangle pqst with a semicircular region at each end. PQ = 200 metres PT = 70 metres.

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