Question

Show that of all rectangles with a fixed perimeter, the square has the greatest area.

Hint:

To maximize or minimize a quadratic function, take the derivative, set it to zero, and solve for the variable.

Answer 1

Let the length and width of the rectangle be denoted by \( l \) and \( w \) respectively. Given a fixed perimeter \( P \), we have: \[ 2l + 2w = P \quad \Rightarrow \quad l + w = \frac{P}{2} \] The area \( A \) of the rectangle is calculated as: \[ A = l \times w \] Express \( w \) in terms of \( l \): \[ w = \frac{P}{2} - l \] Substitute this expression for \( w \) into the area equation: \[ A = l \left(\frac{P}{2} - l\right) = \frac{P}{2}l - l^2 \] To find the value of \( l \) that maximizes \( A \), we treat \( A \) as a quadratic function of \( l \) and find its derivative with respect to \( l \): \[ \frac{dA}{dl} = \frac{P}{2} - 2l \] Setting the derivative to zero to find the critical point: \[ \frac{P}{2} - 2l = 0 \quad \Rightarrow \quad l = \frac{P}{4} \] Using the relationship \( w = \frac{P}{2} - l \), we find: \[ w = \frac{P}{4} \] This implies that the length and width are equal, meaning the rectangle is a square. Therefore, for a constant perimeter, the square encloses the maximum possible area.