# A track and field playing area is in the shape of a rectangle with semicircles at each end. The inside perimeter of the track is to be 1500 meters. What should the dimensions of the rectangle be so that the area of the rectangle is a maximum?

Jul 1, 2016

The rectangle with maximum area should have dimensions
$L = 375$ (the longer sides)
$W = \frac{750}{\pi} \approx 238.73$ (the shorter sides with semicircles)

#### Explanation:

Assume the dimensions of a base rectangle are $L$x$W$ with sides of the length $W$ being diameters of semicircles.

Then the perimeter of a track equals to the lengths of two other sides ($2 L$) plus the length of two semicircles (or, equally, one full circle) of a diameter $W$, that $\pi W$.
The total perimeter is, therefore
(a) $2 L + \pi W = 1500$

We have to maximize the area of a rectangle, that is
(b) $A = L \cdot W$.

From equation (a) we can derive
(c) $L = \frac{1500 - \pi W}{2}$
and substitute it into the formula (b) for area of rectangle:
$A = \frac{1500 - \pi W}{2} \cdot W = - \frac{\pi}{2} {W}^{2} + 750 W$

This expression is a quadratic polynomial of $W$ with negative coefficient at ${W}^{2}$. As we know, it reaches its maximum at
$W = - \frac{750}{2 \cdot \left(- \frac{\pi}{2}\right)} = \frac{750}{\pi}$

Having $W$ determined, use equation (c) to determine $L$:
$L = \frac{1500 - \pi \left(\frac{750}{\pi}\right)}{2} = 375$

CHECK against equation (a):
$2 \cdot 375 + \pi \frac{750}{\pi} = 750 + 750 = 1500$