JAMB 2001 · UME · Q37

Find the dimensions of the rectangle of greatest area which has a fixed perimeter $p$ .

Worked solution (try it first)

Let the sides be $x$ and $y$ .
The perimeter is $2x + 2y = p$ , so $y = \frac p2 - x$ and the area is $A = x\left(\frac p2 - x\right)$
$= \frac p2x - x^2$ .
At the maximum $\frac{dA}{dx} = \frac p2 - 2x = 0$ , so $x = \frac p4$ .
Then $y = \frac p2 - \frac p4 = \frac p4$ too: a square of sides $\frac p4$ , option A.