The Rectangle of Greatest Area with a Given Perimeter

In the setting of Single-Variable Calculus on an Interval, let $P\in\mathbb{R}$ with $0<P$, and let $[0,P/2]$ be the closed interval determined by $0$ and $P/2$. Define $A:[0,P/2]\to\mathbb{R}$ by

$$A(x)=x\left(\frac{P}{2}-x\right).$$

For $x\in[0,P/2]$ the numbers $x$ and $P/2-x$ are both nonnegative and are the side lengths of a rectangle of perimeter $2\bigl(x+(P/2-x)\bigr)=P$, whose area is $A(x)$; and every rectangle of perimeter $P$ arises this way: if such a rectangle has adjacent sides of lengths $x$ and $y$, then $2(x+y)=P$, so $y=P/2-x$, and $0\le x\le P/2$ because $x$ and $y$ are nonnegative.

Problem. Show that $A$ attains a greatest value on $[0,P/2]$, and that the only point of $[0,P/2]$ at which this greatest value is attained is $x=P/4$. Since the two side lengths are then both equal to $P/4$, this says that among all rectangles of perimeter $P$ the square has the greatest area.

The intended route combines Extreme Value Theorem on a Closed Real Interval with Vanishing of the Derivative at a Local Extremum Strictly Inside an Interval.