Chapter 3.4, Problem 17AYU

Constructing a Stadium A track-and-field playing area is in the shape of a rectangle with semicircles at each end. See the figure (top, right).The inside perimeter of the track is to the 1500 meters. What should the dimensions of the rectangle be so that the area of the rectangle is a maximum?

To determine

To calculate: The dimension of the rectangle so that the area of the rectangle is maximum.

Answer to Problem 17AYU

Solution:

The rectangle is of the dimension $238.73$ meters by 375 meters.

Explanation of Solution

Given:

The inside perimeter of the track is 1500 meters.

The shape of the field is

Formula used:

The area of the rectangle is the product of its length and breadth.

Perimeter of the whole track is the perimeter of the rectangular track plus the perimeter of both the semi circles minus both the breadth sides of the rectangle.

Therefore, we have $P=2\pi r+2x$

Calculation:

Here, we know that the inside perimeter is 1500 meters.

The radius of the semi circle is $r=\frac{y}{2}$

Therefore, the perimeter of the field is

$1500=2\pi \left(\frac{y}{2}\right)+2x$

$\Rightarrow 1500=2x+\pi y$

Using the above equation, we get

$1500=2x+\pi y$

$\Rightarrow x=\frac{1500-\pi y}{2}$

$\Rightarrow x=750-\frac{\pi }{2}y$

Now, we have to find the dimensions of the rectangle for maximum area.

Therefore, the area of the rectangle is

Now, we can see that the area is a quadratic function with $a=-\frac{\pi }{2},b=750,c=0$.

Since,$a$ is negative, the maximum value is found at the vertex.

Thus, the point of the maximum value is

$y=-\frac{b}{2a}=-\frac{750}{2\left(-\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}$

$=\frac{750}{\pi }=238.73$

Thus, the breadth of the rectangle is $238.73$ meters.

Now, we need to find the length.

Thus, we have

$x=750-\frac{\pi }{2}(238.73)$

$=375$

Thus, the length of the rectangle is 375 meters.