Arc length contest! Minimize the arc length of $f(x)$ when given three conditions. Contest: Give an example of a continuous function $f$ that satisfies three conditions:


*

*$f(x) \geq 0$ on the interval $0\leq x\leq 1$;

*$f(0)=0$ and $f(1)=0$;

*the area bounded by the graph of $f$ and the $x$-axis between $x=0$ and $x=1$ is equal to $1$. 


Compute the arc length, $L$, for the function $f$. The goal is to minimize $L$ given the three conditions above. 

$\mathbf{\color{red}{\text{Contest results:}}}$ 
$$
\begin{array}{c|ll}
\hline 
\text{Rank} & \text{User} & {} & {} & \text{Arc length} \\ \hline
\text{1} & \text{robjohn $\blacklozenge$} & {} & {} & 2.78540 \\ 
\text{2} & \text{Glen O} & {} & {} & 2.78567 \\ 
\text{3} & \text{mickep} & {} & {} & 2.81108 \\ 
\text{4} & \text{mstrkrft} & {} & {} & 2.91946 \\ 
\text{5} & \text{MathNoob} & {} & {} & 3.00000 \\\hline 
\text{-} & \text{xanthousphoenix} & {} & {} & 2.78540 \\ 
\text{-} & \text{Narasimham} & {} & {} & 2.78 \\ 
\end{array}$$

Original question after contest statement: The contest question was inspired by this paper. Can anyone come up with a different entry than those listed in the table below?
$$
\begin{array}{c|ll}
\hline 
\text{Rank} & \text{Function} & {} & {} & \text{Arc length} \\ \hline
\text{1} & 1.10278[\sin(\pi x)]^{0.153764} & {} & {} & 2.78946 \\ 
\text{2} & (8/\pi)\sqrt{x-x^2} & {} & {} & 2.91902 \\ 
\text{3} & 1.716209468\sqrt{x}\,\mathrm{arccos}(x) & {} & {} & 2.91913 \\ 
\text{4} & (8/\pi)x\,\mathrm{arccos}(x) & {} & {} & 3.15180 \\ 
\text{5} & (15/4)x\sqrt{1-x} & {} & {} & 3.17617 \\ 
\text{6} & -4x\ln x & {} & {} & 3.21360 \\ 
\text{7} & 10x(1-\sqrt{x}) & {} & {} & 3.22108 \\ 
\text{8} & -6x^2+6x & {} & {} & 3.24903 \\ 
\text{9} & 9.1440276(2^x-x^2-1) & {} & {} & 3.25382 \\ 
\text{10} & (-12/5)(x^3+x^2-2x) & {} & {} & 3.27402 \\ 
\end{array}$$

Surprised this is still an open problem.
This is the classical Dido's isoperimetric problem in Calculus of Variations. Same is the circle segment solution whether area is to be maximized for given perimeter length or for a given area the perimeter is to be minimized. We compute in the major circle segment.
If $R$ is the radius and $a =0.5$ the semi-chord length, then area
$$ A =1= a \sqrt {R^2-a^2} +R^2 (\pi-\sin^{-1} (a/R)) $$
By numerical iteration ( Newton-Raphson etc. )
$$ R= 0.6131365125223184 $$
and perimeter of major arc
$$ 2 R (\pi - \sin^{-1}(a/R)) = 2.6831297778598477 $$
coinciding with robjohn's earlier solution. The part of his answer after "Problem" is ..
