Largest Rectangle with Given Perimeter is Square

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\SS$ be the set of all rectangles with a given perimeter $L$.

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


Proof

Consider an arbitrary element $R$ of $\SS$.

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 $\AA$ of $R$ is then given by:

$\AA = x \paren {B - x}$

Let $\AA$ be expressed in functional notation as:

$\map f x = x \paren {B - x}$

We have:

\(\ds \map f {x + E} - \map f x\) \(=\) \(\ds \paren {x + E} \paren {B - \paren {x + E} } - x \paren {B - x}\)
\(\ds \) \(=\) \(\ds E B - 2 E x - E^2\)
\(\ds \leadsto \ \ \) \(\ds \frac {\map f {x + E} - \map f x} E\) \(=\) \(\ds B - 2 x - E\)
\(\ds \leadsto \ \ \) \(\ds \intlimits {\frac {\map f {x + E} - \map f x} E} {E \mathop = 0} {}\) \(=\) \(\ds B - 2 x\)


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.

$\blacksquare$


Historical Note

The result Largest Rectangle with Given Perimeter is Square was given by Pierre de Fermat as one of the first tests to demonstrate the use of his result Derivative at Maximum or Minimum.

It is a more specific (and accessible) version of Proposition $27$: Similar Parallelogram on Half a Straight Line.


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Largest_Rectangle_with_Given_Perimeter_is_Square&oldid=448984"