Question

A farmer is building a fence to enclose a rectangular area consisting of two separate regions. The four walls and one additional vertical segment (to separate the regions) are made up of fencing, as shown below.A rectangular area consisting of two separated regions.A rectangular area consisting of two separated regions.If the farmer has 246 feet of fencing, what are the dimensions of the region which enclose the maximal area?

Answer 1

Let the width be x and the length be y as shown in the figure.

The perimeter of the fence is 246 feet so it follows:

$\begin{gathered} 3x+2y=246 \\ y=\frac{246-3x}{2} \end{gathered}$

The area of the fence is given by:

$\begin{gathered} A=xy \\ A=x(\frac{246-3x}{2}) \\ A=123x-\frac{3}{2}x^2 \end{gathered}$

Differentiate w. r. t. x to get:

$\frac{dA}{dx}=123-3x$

For maxima and minima calculation, the derivative is zero so it follows:

$\begin{gathered} 123-3x=0 \\ 3x=123 \\ x=\frac{123}{3}=41 \end{gathered}$

So the width is 41 feet and the length y is given by:

$y=\frac{246-3(41)}{2}=\frac{123}{2}=61.5$

So the width is 41 feet and length is 61.5 feet, the area covered is 2521`.5 sq feet.