= Solution
Write the speed as $v=\sqrt{\dot x^2+\dot y^2}>0$ and take
$$
\mathcal L=\frac12(x\dot y-y\dot x)+\lambda v.
$$
The two <Euler Lagrange equations> are
$$
\boxed{\dot y-\lambda\frac d{dt}\left(\frac{\dot x}{v}\right)=0,\qquad
-\dot x-\lambda\frac d{dt}\left(\frac{\dot y}{v}\right)=0.}
$$
For 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 gives
$$
\boxed{(x-c_2)^2+(y+c_1)^2=\lambda^2.}
$$
Thus 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.
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