Why does a circle enclose the largest area?

Here is a physicist's answer:

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

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.