Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2009 / Paper 1 / 10

At a regular constrained extremum, the method of Lagrange multipliers makes $I+\lambda J$ stationary for a constant $\lambda$. For an endpoint-vanishing variation $\eta$, integration by parts givesThe fundamental lemma of the calculus of variations therefore yieldsThis Euler-Lagrange equation must be supplemented by the two endpoint conditions and the prescribed integral constraint, which determine the constants and multiplier. Existence of the normal multiplier assumes that the first variation of $J$ is not identically zero. If that regularity fails, the general multiplier equation uses $\alpha I+\lambda J$, with $(\alpha ,\lambda )\ne (0,0)$; one cannot automatically set $\alpha =1$.

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.