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

Let $\displaystyle \mathbf{P}$ be the perimeter of the rectangle, which is a constant. Also assume ' $\displaystyle \mathbf{x}$ ' and ' $\displaystyle \mathbf{y}$ ' be the length and breadth of the rectangle, then \[2(x+y)=P \text { and } A(\text { Area })=x y=\frac{x}{2}(P-2 x)=\frac{1}{2}\left(P x-2 x^{2}\right) \] $\displaystyle A^{\prime \prime}(x)=-2<0$ at $\displaystyle x=\frac{P}{4}, \therefore$ Area of the rectangle is max. if it is a square.