= Solution
At a constrained <stationary point> where $\nabla g\ne0$, every tangent direction to $g=0$ is orthogonal to $\nabla g$. Vanishing of the directional derivative of $f$ in all such directions forces $\nabla f=\lambda\nabla g$ for a <Lagrange multiplier> $\lambda$. Solve these equations together with the <constraint>, then check the candidates and any singular or limiting cases for minima or maxima.
Let a closed cylinder have radius $r>0$ and height $h>0$, with fixed $V>0$. Its area is $A=2\pi r^2+2\pi rh$ and the <constraint> is $\pi r^2h=V$. The <Lagrange multiplier> equations give
$$
4\pi r+2\pi h=\lambda\,2\pi rh,\qquad2\pi r=\lambda\pi r^2.
$$
The second yields $\lambda=2/r$; substituting into the first gives $h=2r$. Thus the <minimum-area closed cylinder at fixed volume> has
$$
\boxed{r=\left(\frac V{2\pi}\right)^{1/3},\qquad h=2\left(\frac V{2\pi}\right)^{1/3},\qquad A_{\min}=3(2\pi)^{1/3}V^{2/3}.}
$$
To prove it is the <global minimum>, eliminate $h$ to obtain $A(r)=2\pi r^2+2V/r$. Its derivative is $4\pi r-2V/r^2$, negative before the displayed radius and positive afterwards; the area tends to infinity both as $r\downarrow0$ and as $r\to\infty$. This establishes uniqueness and rules out limiting smaller areas.
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