= Solution
The functional is unbounded above. Let $\ell$ be the positive affine interpolant of the endpoint values and choose a nonnegative, nonzero, continuously differentiable $h$ vanishing at both endpoints. The admissible functions $y_M=\ell+Mh$ satisfy
$$
J[y_M]\ge2\pi\int_{x_1}^{x_2}y_M\,dx\longrightarrow\infty.
$$
This excludes a global maximum, but \b[does not prove that a stationary curve is a minimum]. The concluding assertion is false without additional conditions.
An explicit counterexample uses $x_1=-2$, $x_2=2$, $y_1=y_2=\cosh2$ and the stationary function $y=\cosh x$. Its <second variation>, for a perturbation $h$ vanishing at the endpoints, is
$$
\frac{\delta^2J[h]}{2\pi}
=\int_{-2}^{2}\bigl(2\tanh x\,hh'+\operatorname{sech}^2x\,h'^2\bigr)dx
=\int_{-2}^{2}\operatorname{sech}^2x\,(h'^2-h^2)\,dx.
$$
Choose $h=1-x^2/4$. Since $\operatorname{sech}^2x\le4e^{-2|x|}$, the positive contribution is at most $\int_{\mathbb R}x^2e^{-2|x|}dx=1/2$. On $[-1,1]$, $h^2\ge9/16$, so the magnitude of the negative contribution is at least $(9/8)\tanh1>9/16>1/2$. Thus $\delta^2J[h]<0$. Small perturbations in this direction preserve positivity and the endpoint data while decreasing <surface area>. Consequently this <catenoid> is not a local minimum. The valid conclusion from part (ii) is a necessary stationary form; minimization requires a separate stability and existence analysis.
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