Chapter 7.7, Problem 13P

To determine

To find the maximum area of a rectangular field that can be enclosed and separated by a fence.

Answer to Problem 13P

The maximum area of the rectangular field enclosed by a fence is 26666.666 m².

Explanation of Solution

Given information:

A rectangular field is enclosed by a fence and divided into two parts by another fence. The field is enclosed and separated with 800 m of fencing.

Let, the vertical and horizontal length of the fence around a rectangular field are $x$ meters and $y$ meters respectively.

So, the total length of the fence $=\left(3x+2y\right)$ m.

According to the question,

  $\begin{array}{l}3x+2y=800\\ 2y=800-3x\\ y=\frac{800-3x}{2}=400-1.5x\end{array}$

So, the horizontal length of the fence is $\left(400-1.5x\right)$ m.

According to the given figure,

The horizontal length of the fence $=$ The length of the rectangular field.

The vertical length of the fence $=$ The width of the rectangular field.

  $\therefore$ The area of the rectangular field enclosed by a fence,

  $\text{A}$$=x\cdot y=x\cdot \left(400-1.5x\right)=\left(400x-1.5x^2\right)$ m².

The standard form of a quadratic equation: $y=a{x}^2+bx+c$ .

So, the formula of the axis of symmetry: $x=\frac{-b}{2a}$ .

Now, comparing the area of the rectangular field with the standard form of the quadratic equation:

  $a=-1.5,b=400\text{ and }c=0$ .

By applying the formula of the axis of symmetry:

  $x=\frac{-b}{2a}=\frac{-400}{2\cdot (-1.5)}=\frac{200}{1.5}\approx 133.333$ .

  $\therefore$ The width of the rectangular field enclosed by a fence is 133.333 m.

  $\therefore$ The maximum area of the rectangular field enclosed by a fence,

  $\text{A}$$=400x-1.5x^2=400\times 133.333-1.5\times {\left(133.333\right)}^2\approx 26666.666$ m².

Hence,

The maximum area of the rectangular field enclosed by a fence is 26666.666 m².