Maximum-area rectangles Of all rectangles with a fixed perimeter of P, which one has the maximum area? (Give the dimensions in terms of P.)

Verified step by step guidance

1. Start by expressing the perimeter of the rectangle in terms of its length (L) and width (W). The perimeter P is given by the formula: \( P = 2L + 2W \).
2. Solve the perimeter equation for one of the variables, say W, in terms of L and P: \( W = \frac{P}{2} - L \).
3. Express the area A of the rectangle in terms of L and W. The area is given by: \( A = L \times W \). Substitute the expression for W from the previous step: \( A = L \left( \frac{P}{2} - L \right) \).
4. Simplify the expression for the area: \( A = \frac{P}{2}L - L^2 \). This is a quadratic function in terms of L, which can be written as \( A = -L^2 + \frac{P}{2}L \).
5. To find the maximum area, determine the vertex of the quadratic function \( A = -L^2 + \frac{P}{2}L \). The vertex form of a quadratic \( ax^2 + bx + c \) has its maximum (or minimum) at \( x = -\frac{b}{2a} \). Here, \( a = -1 \) and \( b = \frac{P}{2} \), so the length L that maximizes the area is \( L = \frac{P}{4} \). Substitute back to find W: \( W = \frac{P}{4} \). Thus, the rectangle with maximum area is a square with side length \( \frac{P}{4} \).