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The area of a rectangle is one of the most fundamental concepts in geometry. Understanding how to compute it helps in solving many mathematical problems. The formula for the area of a rectangle (\( A \)) is \( A = L \times W \), where \( L \) is the length and \( W \) is the width. These two dimensions form the basis of the rectangle's shape.
In the context of this exercise, we're looking at rectangles with a constant perimeter. The perimeter (\( P \)) of a rectangle is calculated by adding twice the length and twice the width, \( P = 2L + 2W \).
By re-arranging the perimeter formula, we can express the width in terms of length:

• \( W = \frac{P - 2L}{2} \)

This means that any change in length results in a change in width if the perimeter is kept constant.
Maximizing the area involves adjusting both the length and the width such that their product, \( L \times W \), reaches its greatest value.
Interestingly, among all rectangles with the same perimeter, the rectangle that is a square has been proven to have the largest area.

A square is a special type of rectangle where all sides are equal. This simple property leads to many significant implications in geometry. When discussing rectangles with a given perimeter, maximizing the area often leads us toward the properties of a square.
For a square, the formula for the perimeter simplifies to \( P = 4L \) (since \( L = W \)). Solving for \( L \) gives \( L = \frac{P}{4} \). Hence, to understand why a square can maximize the area, consider its area \( A \):

• \( A = L \times L = \frac{P^2}{16} \)

This result shows that squaring both sides yields the maximum product for a given perimeter, compared to any other configuration of length and width. The identical side lengths contribute to the efficiency of space usage—a unique property of squares.

Mathematical proofs provide a solid foundation for understanding and verifying claims in geometry. In this exercise, the proof revolves around comparing areas of rectangles with a fixed perimeter.
To demonstrate that a square has the largest area, we take the general rectangle's area formula, \( A = L \times (\frac{P}{2} - L) \), and analyze its possible values.
By setting \( L = \frac{P}{4} \), we recognize a square, and the area indeed becomes \( \frac{P^2}{16} \). Testing other values for \( L \) (e.g., \( 0 \) or \( \frac{P}{2} \)), results in a zero area, reflecting degenerate (flattened) shapes.
The key part of the proof is recognizing that as \( L \) approaches \( \frac{P}{4} \), the rectangle approaches a square, while the area reaches its maximum possible value, thus confirming the square's superiority in terms of area coverage for a given perimeter.