A result in a tiny abstract

Result. Fix the length of the fence around a rectangular garden, say twenty metres. Among all rectangles with that fence, the square has the largest area.

Result. Fix the fence $P=20$ m around a rectangular garden. Among all rectangles with $2(a+b)=P$, the square has the largest area $A=ab$.

$P=20$. $\{a\times b\mid 2(a+b)=P\}$: $\max A=ab$ at $a=b$.

Why it is believable. With sides five and five the area is twenty-five. With six and four it is twenty-four. With nine and one it is only nine.

Believable: $5\times 5$ gives $A=25$; $6\times 4$ gives $24$; $9\times 1$ gives $9$.

$5\times 5\to 25$. $6\times 4\to 24$. $9\times 1\to 9$.

Both conditions matter. If the fence length is not fixed, the area can be made as large as you like. If the shape need not be a rectangle, a circle does better still.

Both conditions matter. $P$ not fixed: $A$ is unbounded. Shape not a rectangle: a circle gives $\frac{P^2}{4\pi }\approx 31.8$.

$P$ free $\Rightarrow \sup A=\infty$. Not a rectangle: $◯\to \frac{P^2}{4\pi }\approx 31.8$.

In general, for a fence of length p, the square has sides p over four and area p squared over sixteen, and no other rectangle does better.

In general: $a=b=\frac{P}{4}$, $A=\frac{P^2}{16}$, and for every other rectangle $A\le \frac{P^2}{16}$.

$\forall$ rectangles, $2(a+b)=P$: $A\le \frac{P^2}{16}$, equal at $a=b=\frac{P}{4}$.

In one line: for a fixed perimeter, the square wins among rectangles, and only among rectangles.

In one line: $P$ fixed, rectangles only, the square wins.

$P$ fixed, $\square$: $\max A$ at $a=b$.