Question

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 $1500\mathrm{meters}$ .

What should the dimensions of the rectangle be so that the area of the rectangle is a maximum?

Answer 1

Step-1. Find the area of the rectangle:

Let $x$ be the length of the rectangle and $y$ be the width of the rectangle and side $y$ is being diameters of half circles.

Using the formula of the perimeter of the rectangle and the half circle and the total perimeter of the track in below :

$\begin{array}{rcl}2x+\frac{\mathrm{\pi y}}{2}+\frac{\mathrm{\pi y}}{2}& =& 1500\\ \therefore 2x+\mathrm{\pi y}& =& 1500...\left(1\right)\end{array}$

Solve for $y$ :

$y=\frac{1500-2x}{\mathrm{\pi }}...\left(2\right)$

Using the formula of the area of the rectangle in below :

$\begin{array}{rcl}A& =& x\times y\left[\text{Area of rectangle}\right]\\ & \Rightarrow & A=x\times \left(\frac{1500-2x}{\mathrm{\pi }}\right)\left[\because y=\frac{1500-2x}{\mathrm{\pi }}\right]\\ \therefore A& =& \frac{1500x-2x^2}{\mathrm{\pi }}\end{array}$

Step-2: Find the value of $x$ for which the area is maximum:
Since the area is maximum, take derivative with respect to $x$ and set it equal to zero :

$\begin{array}{rcl}\frac{dA}{dx}& =& \frac{d}{dx}\left(\frac{1500x-2x^2}{\mathrm{\pi }}\right)\left[\text{Take derivative}\right]\\ & \Rightarrow & 0=\frac{1500\times 1-4x}{\mathrm{\pi }}\left[\text{Substitute}\frac{dA}{dx}=0\right]\\ & \Rightarrow & 1500-4x=0\left[\text{Simplifying}\right]\\ & \Rightarrow & 4x=1500\\ & \Rightarrow & x=\frac{1500}{4}\\ \therefore x& =& 375\end{array}$

Step-3: Find the value of $y$:

Substitute the value of $x$ in equation $\left(2\right)$ :

$\begin{array}{rcl}y& =& \frac{1500-2\times 375}{\mathrm{\pi }}\left[\text{Simplifying}\right]\\ & \Rightarrow & y=\frac{1500-750}{\mathrm{\pi }}\\ & \Rightarrow & y=\frac{750}{\mathrm{\pi }}\\ \therefore y& \approx & 238.73\end{array}$

Hence, the dimensions of the rectangle are $\mathbf{375}\mathbf{}\mathbf{meters}$ and $\mathbf{238}\mathbf{.}\mathbf{73}\mathbf{}\mathbf{meters}$.