The problem asks to find the length and width of a rectangle that maximizes the area.

1. Problem Description

The problem asks to find the length and width of a rectangle that maximizes the area.

2. Solution Steps

To maximize the area of a rectangle with a fixed perimeter, the rectangle should be a square. This is because for a given perimeter, a square encloses the largest area compared to any other rectangle.

Let $l$ be the length and $w$ be the width of the rectangle. Let $A$ be the area.

$A = l \times w$

If we have a fixed perimeter $P$, then

$P = 2l + 2w$

$w = \frac{P - 2l}{2} = \frac{P}{2} - l$

Substitute $w$ into the area equation:

$A = l(\frac{P}{2} - l) = \frac{P}{2}l - l^2$

To find the maximum area, we can complete the square or take the derivative.

Completing the square:

$A = - (l^2 - \frac{P}{2}l) = - (l^2 - \frac{P}{2}l + (\frac{P}{4})^2 - (\frac{P}{4})^2) = - (l - \frac{P}{4})^2 + (\frac{P}{4})^2$

The maximum area occurs when $l = \frac{P}{4}$.

Then $w = \frac{P}{2} - \frac{P}{4} = \frac{P}{4}$.

Thus $l = w$, which means the rectangle is a square.

Alternatively, we can use calculus.

$\frac{dA}{dl} = \frac{P}{2} - 2l$

Set $\frac{dA}{dl} = 0$ to find critical points:

$\frac{P}{2} - 2l = 0$

$2l = \frac{P}{2}$

$l = \frac{P}{4}$

$w = \frac{P}{2} - \frac{P}{4} = \frac{P}{4}$

So, $l=w$.

Thus, the maximum area for a given perimeter occurs when the rectangle is a square, meaning the length and width are equal.

3. Final Answer

The length and width should be equal (i.e., the rectangle is a square) to produce the largest possible area. This is because, for a fixed perimeter, a square encloses the largest area compared to any other rectangle.