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.

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