Answer:
The length of 20 feet and width of 40 feet will result in the least amount of fencing.
Step-by-step explanation:
Please find the attachment.
Let w represent width and l represent length of the rectangle.
We have been given that a rectangular area against a wall is to be fenced off on the other three sides to enclose 800 square feet.
We know that area of rectangle is width times length that is:
[tex]A=w\cdot l[/tex]
[tex]800=w\cdot l[/tex] This is our constraint equation.
We can see from the attachment that the fencing would be for 3 sides that is:
[tex]\text{Perimeter}=l+l+w[/tex]
[tex]P=2l+w[/tex] This is our objective equation.
From constraint equation, we will get:
[tex]w\cdot l=800[/tex]
[tex]w=\frac{800}{l}[/tex]
Substitute this value in objective equation:
[tex]P=2l+\frac{800}{l}[/tex]
[tex]P=2l+800l^{-1}[/tex]
Let us find the derivative of objective equation.
[tex]P'=2-800l^{-2}[/tex]
Now, we will set the derivative equal to 0 to solve for length:
[tex]2-800l^{-2}=0[/tex]
[tex]2-\frac{800}{l^2}=0[/tex]
[tex]-\frac{800}{l^2}=-2[/tex]
Cross multiply:
[tex]-2l^2=-800[/tex]
[tex]\frac{-2l^2}{-2}=\frac{-800}{-2}[/tex]
[tex]l^2=400[/tex]
Take positive square root:
[tex]l=\sqrt{400}[/tex]
[tex]l=20[/tex]
Upon substituting [tex]l=20[/tex] in [tex]w=\frac{800}{l}[/tex], we will get:
[tex]w=\frac{800}{20}[/tex]
[tex]w=40[/tex]
Therefore, the length of 20 feet and width of 40 feet will result in the least amount of fencing.
