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. The perimeter \( P \) is fixed, so: \[ 2l + 2w = P \quad \Rightarrow \quad l + w = \frac{P}{2} \] The area \( A \) of the rectangle is given by: \[ A = l \times w \] Now, express \( w \) in terms of \( l \): \[ w = \frac{P}{2} - l \] Substitute this into the area equation: \[ A = l \left(\frac{P}{2} - l\right) = \frac{P}{2}l - l^2 \] This is a quadratic function, and we need to find the value of \( l \) that maximizes \( A \). To do this, we take the derivative of \( A \) with respect to \( l \): \[ \frac{dA}{dl} = \frac{P}{2} - 2l \] Set \( \frac{dA}{dl} = 0 \) to find the critical point: \[ \frac{P}{2} - 2l = 0 \quad \Rightarrow \quad l = \frac{P}{4} \] Since \( w = \frac{P}{2} - l \), we also have: \[ w = \frac{P}{4} \] Thus, the length and width are equal, and the rectangle is a square. This shows that for a fixed perimeter, the square has the greatest area.