Chapter 3, Problem 24RE

24. Enclosing the Most Area with a Fence A farmer with 10,000 meters of fencing wants to enclose a rectangular field and then divide it into two plots with a fence parallel to one of the sides. See the figure. What is the largest area that can be enclosed?

To determine

To calculate: The maximum area that can be enclosed with a fence of length 10000 meters.

Answer to Problem 24RE

Solution:

The maximum area enclosed by the fence will be $4166683.33$ square meters.

Explanation of Solution

Given:

The farmer wants to enclose the rectangular plot with the fencing of length 10000 meters and then divide the plot into 2 with fencing parallel to one of the sides.

Formula used:

Area of a rectangle is $Area=length\times width$

Perimeter of a rectangle is $P=2(l+w)$, where $l$ is the length and $w$ is the width.

For a quadratic function $f(x)=a{x}^2+bx+c$, the vertex $\left(\frac{-b}{2a},f\left(\frac{-b}{2a}\right)\right)$ is the maximum point if $a$ is negative and the vertex is the minimum point if $a$ is positive.

Calculation:

From the given figure, we can see that the perimeter of the rectangular fencing is $2l+3w=10000$.

We counted width 3 times as there is a dividing fence and 2 sides of the rectangle.

Thus, the equation of the length of the rectangle is

$l=\frac{10000-3w}{2}$

The area of the plot is

$Area=l\times w$

$=\left(\frac{10000-3w}{2}\right)w$

$=-\frac{3}{2}{w}^2+5000w$

Thus, we can see that the area is a quadratic function with $a=-1.5$, which is negative.

Therefore, the vertex if the maximum point.

Thus, we get

$w=\frac{-b}{2a}=\frac{-5000}{2(-1.5)}=1666.67$

Thus, the maximum area will be

$A(1666.67)=-\frac{3}{2}{(1666.67)}^2+5000(1666.67)=4166683.33$

Thus, the maximum area enclosed by the fence will be $4166683.33$ square meters.