Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2001 / Paper 2 / 6

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.

Maximize the total surface area $A=2\pi r(r+h)$ with the constraint $g=r^2+h^2/4=1$. The Lagrange multiplier equations areFrom the second, $\lambda =4\pi r/h$. Substitute into the first and multiply by $h/(2\pi )$ to obtain $2rh+h^2=4r^2$. Thus the aspect ratio $q=h/r$ satisfies $q^2+2q-4=0$. Since $q>0$,This is the maximum-area cylinder inscribed in a sphere. The dimensions and area, although not needed to specify the ratio, areThere is only one positive stationary ratio. At the degenerate boundaries $r=0$ and $h=0$, the limiting areas are zero and $2\pi$, both below this value, confirming the maximum.