Isoperimetric Inequality

Theorem

Let $F$ be a plane figure.

Let $P$ be the length of the perimeter of $F$.

Let $A$ be the area of $F$.

Then:

$P^2 \ge 4 \pi A$

where the equality happens when $F$ is a circle.

That is, the circle is the largest of all the isoperimetric figures whose perimeters are of length $P$.

Proof

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Also see

• Dido's Problem

Historical Note

The problem of determining the plane figure which has the maximum area for a given perimeter was first solved by Jacob Bernoulli.

He also went ahead with a generalisation of the problem.

Sources

• 1937: Eric Temple Bell: Men of Mathematics ... (previous) ... (next): Chapter $\text{VIII}$: Nature or Nurture?
• 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): isoperimetric inequality
• 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): isoperimetric inequality