Largest Rectangle with Given Perimeter is Square

Theorem
Let $\mathcal S$ be the set of all rectangles with a given perimeter $L$.

The element of $\mathcal S$ with the largest area is the square with length of side $\dfrac L 4$.

Proof
Consider an arbitrary element $R$ of $\mathcal S$.

Let $B$ be half the perimeter of $R$.

Let $x$ be the length of one side of $R$.

Then the length of an adjacent side is $B - x$.

The area $\mathcal A$ of $R$ is then given by:
 * $\mathcal A = x \left({B - x}\right)$

Let $\mathcal A$ be expressed in functional notation as:
 * $f \left({x}\right) = x \left({B - x}\right)$

We have:

Thus from Derivative at Maximum or Minimum, when $B - 2 x = 0$, the area of $R$ is at a maximum or a minimum.

That is:
 * $x = \dfrac B 2 = \dfrac L 4$

Geometrical analysis shows that for this length of side the area is not a minimum because that happens when $x = 0$ or $x = B$.

Hence the result.

Historical Note
This result was used by to demonstrate the use of his result Derivative at Maximum or Minimum.

It is a more specific (and accessible) version of.