A physical fitness room consists of a rectangular region with a semicircle on each end. If the perimeter of the room is to be a 200 meter running track, how do find the dimensions that will make the area of the rectangular regions as large as possible?

May 2, 2018

 { ("length", =50 \ m), ("width", =50/pi \ m ~~ 15.92 \ m) :}

Explanation:

Let us set up the following variables:

 { (x, "length of rectangular region (m)"), (r, "Radius of semi-circle (m)"), (P, "Total perimeter of the room (m"^2")"), (A, "Total Area of the rectangular region (m"^2")") :}

The perimeter of the given room, is that of two identical semicircular ends, and the two rectangular region lengths. Thus:

$P = \frac{1}{2} \left(2 \pi r\right) + \frac{1}{2} \left(2 \pi r\right) + x + x$

$\setminus \setminus \setminus = 2 \pi r + 2 x$

But we are given that the total perimeter is $200 \setminus m$, a constant, therefore:

$2 \pi r + 2 x = 200 \implies r = \frac{100 - x}{\pi}$

Now the total area of the rectangular region is:

$A = \left(x\right) \left(2 r\right)$
$\setminus \setminus \setminus = \left(x\right) 2 \left(\frac{100 - x}{\pi}\right)$
$\setminus \setminus \setminus = \frac{200 x - 2 {x}^{2}}{\pi}$

For maximum or minimum, we seek a critical point of the area function, $A \left(x\right)$ wrt $x$. We therefore calculate the derivative:

$\frac{\mathrm{dA}}{\mathrm{dx}} = \frac{200 - 4 x}{\pi}$

At a critical point, the first derivative vanishes, thus we require:

$f ' \left(x\right) = 0 \implies 200 - 4 x = 0$

$\therefore x = 50 \setminus m$

We can visually confirm from the graph:
graph{(200x-2x^2)/(pi) [-50, 150, -400, 2000]}

With $x = 50 \implies r = \frac{100 - 50}{\pi} = \frac{50}{\pi}$

that the solution is both consistent and corresponds to a maximum, and the dimensions of the sought room are thus:

 { ("length", =x, =50 \ m), ("width", =2r,=50/pi \ m ~~ 15.92 \ m) :}