Question

Mr. Jones' other friend Ms. O'Neill has "x" meters of fencing, what is the maximum area in square meters she can have?

Answer 1

Answer:

$Max\ Area = \frac{x^2}{16}$

Step-by-step explanation:

Given

$P = x$ ---- the perimeter of fencing

Required

The maximum area

Let

$L \to Length$

$W \to Width$

So, we have:

$P = 2(L + W)$

This gives:

$2(L + W) = x$

Divide by 2

$L + W = \frac{x}{2}$

Make L the subject

$L = \frac{x}{2} - W$

The area (A) of the fence is:

$A = L * W$

Substitute $L = \frac{x}{2} - W$

$A = (\frac{x}{2} - W) * W$

Open bracket

$A = \frac{x}{2}W - W^2$

Differentiate with respect to W

$A' = \frac{x}{2} - 2W$

Set to 0

$\frac{x}{2} - 2W = 0$

Solve for 2W

$2W = \frac{x}{2}$

Solve for W

$W = \frac{x}{4}$

Recall that:

$L = \frac{x}{2} - W$

$L = \frac{x}{2} - \frac{x}{4}$

$L = \frac{2x- x}{4}$

$L = \frac{x}{4}$

So, the maximum area is:

$A = L * W$

$A = \frac{x}{4}*\frac{x}{4}$

$Max\ Area = \frac{x^2}{16}$