Answer:
[tex]Max\ Area = \frac{x^2}{16}[/tex]
Step-by-step explanation:
Given
[tex]P = x[/tex] ---- the perimeter of fencing
Required
The maximum area
Let
[tex]L \to Length[/tex]
[tex]W \to Width[/tex]
So, we have:
[tex]P = 2(L + W)[/tex]
This gives:
[tex]2(L + W) = x[/tex]
Divide by 2
[tex]L + W = \frac{x}{2}[/tex]
Make L the subject
[tex]L = \frac{x}{2} - W[/tex]
The area (A) of the fence is:
[tex]A = L * W[/tex]
Substitute [tex]L = \frac{x}{2} - W[/tex]
[tex]A = (\frac{x}{2} - W) * W[/tex]
Open bracket
[tex]A = \frac{x}{2}W - W^2[/tex]
Differentiate with respect to W
[tex]A' = \frac{x}{2} - 2W[/tex]
Set to 0
[tex]\frac{x}{2} - 2W = 0[/tex]
Solve for 2W
[tex]2W = \frac{x}{2}[/tex]
Solve for W
[tex]W = \frac{x}{4}[/tex]
Recall that:
[tex]L = \frac{x}{2} - W[/tex]
[tex]L = \frac{x}{2} - \frac{x}{4}[/tex]
[tex]L = \frac{2x- x}{4}[/tex]
[tex]L = \frac{x}{4}[/tex]
So, the maximum area is:
[tex]A = L * W[/tex]
[tex]A = \frac{x}{4}*\frac{x}{4}[/tex]
[tex]Max\ Area = \frac{x^2}{16}[/tex]
