# Why does a circle enclose the largest area?

In this wikipedia, article http://en.wikipedia.org/wiki/Circle#Area_enclosed its stated that the circle is the closed curve which has the maximum area for a given arc length. First, of all, I would like to see different proofs, for this result. (If there are any elementary ones!)

One, interesting observation, which one can think while seeing this problem, is: How does one propose such type of problem? Does, anyone take all closed curves, and calculate their area to come this conclusion? I don't think that's the right intuition.

Imagine a rope enclosing a two-dimensional gas, with vacuum outside the rope. The gas will expand, pushing the rope to enclose a maximal area at equilibrium.

When the system is at equilibrium, the tension in the rope must be constant, because if there were a tension gradient at some point, there would be a non-zero net force at that point in the direction of the rope, but at equilibrium the net force must be zero in all directions.

The gas exerts a force outward on the rope, so tension must cancel this force. Take a small section of rope, so that it can be thought of as a part of some circle, called the osculating circle. The force on this rope segment due to pressure is $$P l$$, with $$P$$ pressure and $$l$$ the length. The net force due to tension is $$2 T \sin(l/2R)$$, with $$T$$ tension and $$R$$ the radius of the osculating circle.

Because the pressure is the same everywhere, and the force from pressure must be canceled by the force from tension, the net tension force must be the same for any rope segment of the same length. That means the radius of the osculating circle is the same everywhere, so the rope must be a circle.

For a simple experimental demonstration, we replace the gas with a soap film. A soap film will minimize its area, so if we put a loop of string inside a soap film, then break the film inside the string, the remaining film outside the string will pull the string into a circle. image credit: Carboni, Giorgio. "Experiments on Surface Phenomena and Colloids", http://www.funsci.com/fun3_en/exper2/exper2.htm

• It's cool to see. Physicists have intuition and we need that. Jul 6, 2011 at 22:59
• @PatrickDaSilva: Intuitive or not, this answer is wrong. Idealized soap films achieve a minimal energy configuration, but may not achieve a minimum energy configuration. Furthermore, it is terrible to assume (without proof) that the equilibrium state must have a circumference that has a well-defined radius of curvature. Experimenting with soap films is fun and should be encouraged, but sloppy mathematics should not... Aug 15, 2020 at 3:22
• @I0_0I: Nice! So refreshing to see that someone has gotten my bubbly pun. =P Jul 1, 2021 at 17:13

As Qiaochu Yuan pointed out, this is a consequence of the isoperimetric inequality that relates the length $L$ and the area $A$ for any closed curve $C$:

$$4\pi A \leq L^2 \ .$$

Taking a circumference of radius $r$ such that $2\pi r = L$, you obtain

$$A \leq \frac{L^2}{4\pi} = \frac{4 \pi^2 r^2}{4\pi} = \pi r^2 \ .$$

That is, the area $A$ enclosed by the curve $C$ is smaller than the area enclosed by the circumference of the same length.

As for the proof of the isoperimetric inequality, here is the one I've learnt as undergraduate, which is elementary and beautiful, I think.

Go round your curve $C$ counterclockwise. For a plane vector field $(P,Q)$, Green's theorem says

$$\oint_{\partial D}(Pdx + Qdy) = \int_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dxdy\ .$$

Apply it for the vector field $(P,Q) = (-y,x)$ and when $D$ is the region enclosed by your curve $C = \partial D$. You obtain

$$A = \frac{1}{2} \oint_{\partial D} (-ydx + xdy) \ .$$

Now, parametrize $C= \partial D$ with arc length:

$$\gamma : [0,L] \longrightarrow \mathbb{R}^2 \ ,\qquad \gamma (s) = (x(s), y(s)) \ .$$

Taking into account that

$$0= xy \vert_0^L = \int_0^L x'yds + \int_0^L xy'ds \ ,$$

we get

$$A = \int_0^L xy'ds = -\int_0^L x'yds \ .$$

So enough for now with our curve $C$. Let's look for a nice circumference to compare with!

First of all, $[0,L]$ being compact, the function $x: [0,L] \longrightarrow \mathbb{R}$ will have a global maximum and a global minimum. Changing the origin of our parametrization if necessary, me may assume the minimum is attained at $s=0$. Let the maximum be attained at $s=s_0 \in [0,L]$. Let $q = \gamma (0)$ and $p = \gamma (s_0)$. (If there are more than one minimum and more than one maximum, we choose one of each: the ones you prefer.)

Since $x'(0) = x'(s_0) = 0$, we have vertical tangent lines at both points $p,q$ of our curve $C$. Draw a circumference between these parallel lines, tangent to both of them (a little far away of $C$ to avoid making a mess). So the radius of this circumference will be $r = \frac{\| pq \|}{2}$.

Let's take the origin of coordinates at the center of this circumference. We parametrize it with the same $s$, the arc length of $C$:

$$\sigma (s) = (\overline{x}(s), \overline{y}(s)) \ , \quad s \in [0, L] \ .$$

Of course, $\overline{x}(s)^2 + \overline{y}(s)^2 = r^2$ for all $s$. If we choose $\overline{x}(s) = x(s)$, this forces us to take $\overline{y}(s) = \pm \sqrt{r^2 - \overline{x}(s)^2}$. In order that $\sigma (s)$ goes round all over our circumference counterclockwise too, we choose the minus sign if $0\leq s \leq s_0$ and the plus sign if $s_0 \leq s \leq L$.

We are almost done, just a few computations left.

Let $\overline{A}$ denote the area enclosed by our circumference. So, we have

$$A = \int_0^L xy'ds = \int_0^L \overline{x}y'ds \qquad \text{and} \qquad \overline{A}= \pi r^2 = -\int_0^L\overline{y}\overline{x}'ds = -\int_0^L\overline{y} x'ds \ .$$

Hence,

\begin{align} A + \pi r^2 &= A + \overline{A} = \int_0^L (\overline{x}y' - \overline{y}x')ds \\\ &\leq \int_0^L \vert \overline{x}y' - \overline{y}x'\vert ds \\\ &= \int_0^L \vert (\overline{x}, \overline{y})\cdot (y', -x')\vert ds \\\ &\leq \int_0^L \sqrt{\overline{x}^2 + \overline{y}^2} \cdot \sqrt{(y')^2+ (-x')^2}ds \\\ &= \int_0^L rds = rL \ . \end{align}

The last inequality is Cauchy-Schwarz's one and the last but one equality is due to the fact that $s$ is the arc-length of $C$.

Summing up:

$$A + \pi r^2 \leq rL \ .$$

Now, since the geometric mean is always smaller than the arithmetic one,

$$\sqrt{A\pi r^2} \leq \frac{A + \pi r^2}{2} \leq \frac{rL}{2} \ .$$

Thus

$$A \pi r^2 \leq \frac{r^2L^2}{4} \qquad \Longrightarrow \qquad 4\pi A \leq L^2 \ .$$

There are relatively simple proofs in textbooks on calculus of variations.

In more elementary approaches a convex figure is deformed, in discrete steps or through a continuous unbending process, toward a circle, and two things need to be proved: convergence to the circle, and increase of the isoperimetric ratio throughout the flow. Usually one step is easy and the other is difficult, requiring non-elementary methods to make rigorous. It is also necessary to make explicit what class of curves is considered: rectifiable, piecewise smooth, or something else.

The simplest argument I know that is elementary and rigorous is to prove the finite-dimensional approximation, that for fixed edge lengths of a polygon, there is a maximum area (by compactness) and (by elementary geometry or Lagrange multipliers) it is the one where all vertices are on a circle. Then, use this to prove that any smooth curve, if it beats the circle, has a finite polygonal approximation that beats the inscribed polygon.

As for the second question, the result is quite intuitive. First of all one can readily see that we can suppose WLOG that the curve is convex, so already it cannot be too far from being circle-like. Second, it's not hard to see that stretching the curve out so as to make it non-uniform causes it to enclose less area, e.g. consider the analogous problem for rectangles (or even try it for $n$-gons). The intuition here is that a kink in the boundary of an area does not create much extra area but creates extra arc length.

Of course the hard part is to justify these intuitions. Probably the proof that comes closest proceeds via Steiner symmetrization; there is a link at the Wikipedia article. There is also a neat Fourier-analytic proof.

Here is the Fourier series proof, due to Hurwitz.

Let $f:[0,2\pi]\to\mathbb C$ be a $C^1$ curve parametrized by the arc length. Denote again by $f$ the corresponding $2\pi$-periodic function, and let $c_n$ be its $n$-th Fourier coefficient.

By Stokes's and Parseval's Formulas, the signed area enclosed is $$\frac{1}{2i}\ \int_0^{2\pi}\ f'(t)\ \overline{f(t)}\ dt =\frac{\pi}{i}\ \sum_{n\in\mathbb Z}\ in\ |c_n|^2 =\pi\ \sum_{n\in\mathbb Z}\ n\ |c_n|^2 \le\pi\ \sum_{n\in\mathbb Z}\ n^2\ |c_n|^2=\pi,$$ with equality if and only if $c_n=0$ for $n\not=0,1$, and $|c_1|=1$.

• Could you explain the last equality $\sum_{n\in\mathbb Z}\ n^2\ |c_n|^2=1$?
– caub
May 26, 2014 at 2:23
• @kwak - Dear kwak: The equality follows from the assumption that the curve is parametrized by the arc length. Thanks for your interest. May 26, 2014 at 3:48
• merci @Pierre-Yves it's clear
– caub
May 27, 2014 at 16:28

First you can propose this problem as: The area $A$ encompassed by any simple closed rectifiable curve $C$, of length $L$, satisfacts the inequality $A\geq \frac{L^2}{4\pi}$, and equality occurs, if and only if, $C$ is a circle.

The only proof I have done for this was using Parseval's identity (and therefore Fourier series), so it's not elementary (but it's rather simple if you know the aforementioned identity). Thought if you want I can post that proof.

May I use an intuitive answer here because it may be easy enough for primary school students to think about.

Let's imagine a wire which is bent into two sides. If we want to "enclose" an area as large as possible, any two adjacent sides must be as far away from each other as possible. (Compare with the "enclosed area" when the two sides are very close to each other.) That means the angle between two adjacent sides must tend to 180 degree. But since the length of the wire is fixed and the area need to be enclosed, each side should be as short as possible in order to get larger adjacent angles.

Eventually the shape will become a circle.

I will try to add "Easier" solution yet it requires 2 assumptions:

1. The shape is formed of edges.
2. Given a shape formed of $n$ edges and perimeter $P$ the largest area will be if all edges are equal in their length.

If we take those 2 assumptions something nice happens.
Since the shape is symmetric it has a center and the distance from its center to any vertex is equal.
Let's define the distance to each vertex as $r$.
What is formed is Isosceles Triangle composed from two edges of the length $r$ and a base of the length $\frac{P}{n}$.
The area of this triangle is given by ${A}_{n} = \frac{r}{2} \cos \left( \frac{\pi}{n} \right) \frac{P}{n}$.

Since there are $n$ triangles just like this the area is given by $A = \frac{r}{2} \cos \left( \frac{\pi}{n} \right)$.

Now, to maximize the are one need to maximize the term $\cos \left( \frac{\pi}{n} \right)$ which is maximized for $n \to \infty$ which suggests a shape of a circle.

Dedicated with love to Renana!

Here's a very hand-wavy construction which provides some intuition behind this result:

Begin with a closed convex curve of some perimeter $$P$$. Pick any point on this curve. Now pick find the unique point on this curve that's $$P/2$$ away from this starting point. Bisect the curve along the line defined by these two points. Now look at the two halves. One of them will likely have the larger area. Flip this half over the smaller half, obtaining a new closed curve with larger area and with bilateral symmetry.

Repeated application of this procedure preserves the perimeter, increases the area, and also introduces a new axis of bilateral symmetry with each step. In the limit, the figure obtained is "infinitely bisymmetrical"–a circle. (👋)