Q. 17 - Precalculus Enhanced with Graphing Utilities

Question

Constructing a Stadium A track and field playing area is in the shape of a rectangle with semicircles at each end. See the figure. 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? 

Step-by-Step Solution

Answer

The area of a rectangle is maximized at $L=375m, w=238.7m.$

Step 1. Given information

It is given that the inside perimeter of the truck is to be $1500$ meters. We need to determine the dimensions of the rectangle be so that the area of rectangle is maximum.

Step 2. Simplify

The figure, $r$ represents the radius, $l$ represents the length and $w$ represents the width.

The perimeter of the figure,

$P=2l+2\left(\frac{2\mathrm{\pi r}}{2}\right).$

    $=2l+2\mathrm{\pi r}\dots \left(1\right).$

We know, $w=2r.$

$P=2l+\mathrm{\pi w}.$

The area of the rectangle is given by the equation.

$A=lw\dots \left(2\right).$

We know, $P=1500$ in (1).

$1500=2l+\mathrm{\pi w}.$

$2l=1500-\mathrm{\pi w}.$

$l=750-\frac{\mathrm{\pi w}}{2}.$

Substitute $l$ in (2).

$A=\left(750-\frac{\mathrm{\pi w}}{2}\right)w.$

    $=750w-\frac{{\mathrm{\pi w}}^2}{2}.$

   $=\frac{-{\mathrm{\pi w}}^2}{2}+750w.$

The $w$ coordinate of the vertex of an equation in the form $a{w}^2-bw+c.$

$w=-\frac{b}{2a}.$

    $w=-\frac{750}{2\left(-{\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}.$

        $=\approx 238.7 m.$

Substitute $w$ in equaion 3.

$l=750-\frac{\mathrm{\pi }\left(238.7\right)}{2}.$

   $=375 m.$