 Solution

Write the speed as $v=\sqrt{{\dot{x}}^2+{\dot{y}}^2}>0$ and takeThe two Euler-Lagrange equations areFor example ${\mathcal{L}}_x=\dot{y}/2$ and ${\mathcal{L}}_{\dot{x}}=-y/2+\lambda \dot{x}/v$, yielding the first equation. Integrating once gives constants $c_1,c_2$ with $\lambda \dot{x}/v=y+c_1$ and $\lambda \dot{y}/v=-x+c_2$. The case $\lambda =0$ would force both velocities to vanish, contrary to a regular curve. Squaring and adding therefore givesThus every regular stationary curve is a circle, whose center is set by the integration constants. A simple closed traversal of length $L$ has radius $L/(2\pi )$ and positive enclosed area $L^2/(4\pi )$. The planar isoperimetric inequality proves that this stationary candidate is the global maximum; the necessary variational equations alone would not prove that inequality. The stationary curves of the planar isoperimetric problem give this argument independently of the chosen parametrization.