= Solution
The functional is unbounded above. Let $y_0$ be the positive affine interpolant between the two endpoint values, and choose a smooth nonnegative function $\phi$ that vanishes at the endpoints and is positive somewhere inside. The admissible curves $y_M=y_0+M\phi$ satisfy
$$
J[y_M]\ge2\pi\int y_M\,dx\longrightarrow\infty.
$$
\b[The claimed minimum conclusion does not follow and is false in general.] A stationary point may have a negative <second variation>, even though the functional is unbounded above. Here is an explicit example within the allowed positive-<graph> class.
Take the stationary profile $y(x)=\cosh x$ on $[-2,2]$, and the endpoint-zero variation $\eta(x)=\cosh x-\cosh2$. For sufficiently small perturbation parameter $s$, $y+s\eta$ remains positive and preserves both endpoint values. Put $C=\cosh2$. Differentiating the area twice and integrating its mixed term by parts gives
$$
\frac{\delta^2J[\eta]}{2\pi}=\int_{-2}^2\operatorname{sech}^2x\,(\eta'^2-\eta^2)\,dx
=4C\arctan(\sinh2)-2(1+C^2)\tanh2.
$$
This is strictly negative: the first term is less than $2\pi C$, while $(C+C^{-1})\tanh2>\pi$. For example $e^2>7$ gives $C>7/2$ and $\tanh2>24/25$, whose product already exceeds $\pi$. Thus nearby positive profiles have smaller area. This <catenoid> is \b[not a local minimum].
The corrected conclusion is that any smooth positive minimizer, if attained, must have the <catenoid> form. That necessary condition is not sufficient, and existence or stability requires a separate argument.
\Image[/past-exam-of-the-options-ii-course-of-the-university-of-oxford/2007-catenoid-variation.png]
{title=A catenoid with a negative second variation}
{description=Endpoint-preserving positive perturbations of an extended catenoid lower the area; the stationary profile is not a local minimum.}
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