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At a constrained stationary point where $\nabla g\ne 0$, 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 $\pi r^{2}h=V$. The Lagrange multiplier equations giveThe second yields $\lambda =2/r$; substituting into the first gives $h=2r$. Thus the minimum-area closed cylinder at fixed volume hasTo 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 $r\downarrow 0$ and as $r\to \infty$. This establishes uniqueness and rules out limiting smaller areas.

Use a Lagrange multiplier to maximize $V=\pi r^{2}h$ subject to $2\pi r(r+h)=A$, with $r,h>0$. The equations $\nabla V=\lambda \nabla A$ areThe second equation gives $\lambda =r/2$. Substituting into the first gives $2rh=r(2r+h)$, hence $h=2r$. The constraint then gives $A=6\pi r^2$, soThis is the same aspect ratio as the minimum-area closed cylinder at fixed volume. To confirm maximality directly, eliminate $h$: $V(r)=Ar/2-\pi r^3$ on $0<r<\sqrt{A/(2\pi )}$. Its only stationary point is the value above, its second derivative there is negative, and the volume tends to zero at either end of the interval.