Solution of The Rectangle of Greatest Area with a Given Perimeter

Step 1: $A$ is continuous on $[0,P/2]$. Since $0<P$ we have $0<P/2$. By distributivity,

so $A$ is the restriction to $[0,P/2]$ of the polynomial function on $\mathbb{R}$ with coefficients $c_0=0$, $c_1=P/2$ and $c_2=-1$. By The Real Line: Standing Notation and Background for Calculus §continuity, $A$ is continuous on $[0,P/2]$.

Step 2: a greatest value is attained. Since $0<P/2$ and $A$ is continuous on $[0,P/2]$, Extreme Value Theorem on a Closed Real Interval gives a point $x^{\ast}\in[0,P/2]$ with $A(x)\le A(x^{\ast})$ for every $x\in[0,P/2]$. So a greatest value is attained.

Step 3: every maximiser is an interior point. Let $x^{\ast}\in[0,P/2]$ be any point with $A(x)\le A(x^{\ast})$ for every $x\in[0,P/2]$. We compute

Since $0<P/4$, the product $(P/4)\cdot(P/4)$ is positive by clause 5 of Elementary Order Arithmetic in an Ordered Field, so $0<A(P/4)$. As $0\le P/4\le P/2$, the point $P/4$ lies in $[0,P/2]$, so $0<A(P/4)\le A(x^{\ast})$. Hence $A(x^{\ast})\ne 0$, so $x^{\ast}\ne 0$ and $x^{\ast}\ne P/2$; together with $0\le x^{\ast}\le P/2$ this gives

Step 4: the derivative of $A$. Let $x$ satisfy $0<x<P/2$. Since $0,P/2\in[0,P/2]$, the point $x$ is an interior point of $[0,P/2]$. By clause 1 of Derivative of a Polynomial Function on the Real Line, the restriction to $[0,P/2]$ of $z\mapsto z$ is differentiable at $x$ with derivative $1$, and the restriction of $z\mapsto z^2$ is differentiable at $x$ with derivative $2x$. By clauses 1, 2 and 3 of Sum, Constant Multiple, and Product Rules for One-Dimensional Derivatives, $A$ is differentiable at $x$ with

Step 5: the interior extremum criterion. Let $g$ be the restriction of $A$ to the open interval $(0,P/2)$. Since $0<x^{\ast}<P/2$, the point $x^{\ast}$ lies in $(0,P/2)$, and every point of an open interval is an interior point of it by The Real Line: Standing Notation and Background for Calculus §intervals; so $x^{\ast}$ is an interior point of $(0,P/2)$. As $(0,P/2)$ is an interval contained in $[0,P/2]$, clause 2 of Restriction Stability of Continuity and of the Derivative together with Step 4 shows that $g$ is differentiable at $x^{\ast}$ with $g'(x^{\ast})=A'(x^{\ast})$.

Every $y\in(0,P/2)$ lies in $[0,P/2]$ and therefore satisfies $g(y)=A(y)\le A(x^{\ast})=g(x^{\ast})$. Taking $\delta=1$ in Local Maximum of a Function Relative to a Subset of a Metric Space, the function $g$ has a local maximum at $x^{\ast}$ relative to $(0,P/2)$. By Vanishing of the Derivative at an Interior Local Extremum,

so $2x^{\ast}=P/2$ and hence $x^{\ast}=P/4$.

Conclusion. By Step 2 the greatest value of $A$ on $[0,P/2]$ is attained, and by Steps 3 to 5 every point at which it is attained equals $P/4$. Since a maximiser exists, it is $P/4$, and it is the only one. The corresponding rectangle has side lengths $P/4$ and $P/2-P/4=P/4$, so it is a square.