Q. 24 - Precalculus Enhanced with Graphing Utilities

Question

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? 

Step-by-Step Solution

Answer

The largest area is 4,166,675 m².

Step 1. Given information.

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? 

Step 2. Draw the figure and find the perimeter.

To obtain the largest area that can be obtained we must make use of the equation for the perimeter and area of the field.

Refer to the image below to get a better illustration of the given.

As we can see from the figure the perimeter of the whole field can be obtained through:

$P=3w+2l$

Where l represents the length and w represent the width.

Step 3. Find the area of rectangle.

Since the field is rectangle, then the area of rectangle is obtained by:

$A=lw$

Where l represents the length and w represent the width.

Now obtain the equation for its perimeter. We can obtain a value of l  in term of w.

$$\begin{gathered} P=3w+2l \\ 10000=3w+2l \\ 10000-3w=2l \\ l=\frac{10000}{2}-\frac{3}{2}w \\ l=5000-\frac{3}{2}w \end{gathered}$$

Step 4. Substitute the value of l .

Substitute l to the formula for the area of the field to obtain:

$$\begin{gathered} A=lw \\ A=\left(5000-\frac{3}{2}w\right)w \\ A=5000w-\frac{3}{2}{w}^2 \end{gathered}$$

The x coordinate of the vertex of an equation in the form $a{x}^2+bx+c$, can be obtained using:

$x=-\frac{b}{2a}$

Substitute $a=-\frac{3}{2} \mathrm{and} b=5000$ to equation 2:

$x=-\frac{5000}{2\left(-{\displaystyle \frac{3}{2}}\right)}=1666.67$

Therefore, when the width is approximately 1666.67 meter, the maximum area is obtained.

Step 5. Substitute the value to find l.

$$\begin{gathered} l=5000-\frac{3}{2}\left(1666.67\right) \\ l=5000-2500 \\ l=2500m \end{gathered}$$

Therefore, when the width is 1666.67 m and the length is 2500 m, the maximum area is obtained.

Now substitute the value of l into the area formula to obtain the largest area of the field.

$$\begin{gathered} A=lw \\ A=1666.67\left(2500\right) \\ A=4166675 m^2 \end{gathered}$$

Therefore, the largest area that can be obtained is 4166675 m²