Answer:
Width = 108 ft
Length = 162 ft
Step-by-step explanation:
Define the variables:
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:
[tex]\begin{aligned}\implies 648&=3x+2y\\2y&=648-3x\\y&=324-\dfrac{3}{2}x\end{aligned}[/tex]
The area of the rectangle is the width multiplied by the length.
Create an equation for the area in terms of x:
[tex]\begin{aligned}\implies \textsf{Area}&=\sf width \times length\\A&=xy\\A&=x\left(324-\dfrac{3}{2}x\right)\\A&=324x-\dfrac{3}{2}x^2\end{aligned}[/tex]
To find the value of x when the area is at is maximum, differentiate the equation for A with respect to x.
[tex]\begin{aligned}A&=324x-\dfrac{3}{2}x^2\\\implies \dfrac{\text{d}A}{\text{d}x}&=1 \cdot 324x^{1-1}-2 \cdot \dfrac{3}{2}x^{2-1}\\&=324x^0-3x^1\\&=324(1)-3x\\&=324-3x\end{aligned}[/tex]
Set the differentiated equation to zero and solve for x:
[tex]\begin{aligned}\dfrac{\text{d}A}{\text{d}x}&=0\\\implies 324-3x&=0\\3x&=324\\x&=108 \end{aligned}[/tex]
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:
[tex]\begin{aligned}\implies y&=324-\dfrac{3}{2}(108)\\&=324-162\\&=162\end{aligned}[/tex]
Therefore, the dimensions of the region which enclose the maximal area are:
Differentiation rule used:
[tex]\boxed{\begin{minipage}{4.8 cm}\underline{Differentiating $ax^n$}\\\\If $y=ax^n$, then $\dfrac{\text{d}y}{\text{d}x}=nax^{n-1}$\\\end{minipage}}[/tex]
