Question

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.

If the farmer has 648 feet of fencing, what are the dimensions of the region which enclose the maximal area?

Answer 1

Width = 108 ft

Length = 162 ft

Explanation:

Define the variables:

• Let x = width of the rectangle.
• Let y = length of the rectangle.

If the total length of the fence is the four walls and one additional vertical segment, then the length of the fence is the sum of 3 widths and 2 lengths.

Given the length of the fence is 648 ft, create an expression for the length in terms of the width:

`\begin{aligned}\implies 648&=3x+2y\\2y&=648-3x\\y&=324-(3)/(2)x\end{aligned}`

The area of the rectangle is the width multiplied by the length.

Create an equation for the area in terms of x:

`\begin{aligned}\implies \textsf{Area}&=\sf width * length\\A&=xy\\A&=x\left(324-(3)/(2)x\right)\\A&=324x-(3)/(2)x^2\end{aligned}`

To find the value of x when the area is at is maximum, differentiate the equation for A with respect to x.

`\begin{aligned}A&=324x-(3)/(2)x^2\\\implies \frac{\text{d}A}{\text{d}x}&=1 \cdot 324x^(1-1)-2 \cdot (3)/(2)x^(2-1)\\&=324x^0-3x^1\\&=324(1)-3x\\&=324-3x\end{aligned}`

Set the differentiated equation to zero and solve for x:

`\begin{aligned}\frac{\text{d}A}{\text{d}x}&=0\\\implies 324-3x&=0\\3x&=324\\x&=108 \end{aligned}`

Therefore, the width of the region that encloses the maximal area is 108 ft.

To find the length, substitute the found value of x into the expression for y:

`\begin{aligned}\implies y&=324-(3)/(2)(108)\\&=324-162\\&=162\end{aligned}`

Therefore, the dimensions of the region which enclose the maximal area are:

• Width = 108 ft
• Length = 162 ft

Differentiation rule used:

`\boxed{\begin{minipage}{4.8 cm}\underline{Differentiating $ax^n$}\\\\If $y=ax^n$, then $\frac{\text{d}y}{\text{d}x}=nax^(n-1)$\\\end{minipage}}`