Proof a special extension on the isoperimetric theorem Let both ends of a string of length $L$ be tied to a stick of length $S$. Among all plane regions enclosed by this contraption, it achieves maximum when the string forms a circular arc.
It is noted that if $S=0$, this theorem is reduced to the isoperimetric theorem, which states, for the length $L$ of a closed curve and the area $A$ of the planar region that it encloses, 
$4\pi A \le L^2$,
and that equality holds if and only if the curve is a circle.
Your comments and ideas are greatly appreciated!

By the usual Steiner's reflection argument the solution has to be a convex curve. Moreover, if we take a portion of the curve between two points $A$ and $B$, take the symmetric with respect to the perpendicular bisector of $AB$ and glue that piece back, the new curve has to be convex, too (that transformation does not change the length of the curve nor the enclosed area). Assuming that the solution is a smooth curve, it follows that all the normal vectors have to point towards the same point, so the solution is given by an arc of a circle.
As an alternative, it is not difficult to show through Lagrange's multipliers that among all the convex polygons having side lengths $l_1,l_2,\ldots,l_n$ (in the same order), the cyclic polygon has the maximum area. The solution to the previous problem then follows by taking $l_1=S,l_i=\frac{L}{n-1}$ and letting $n\to +\infty$.
