Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2017 / Paper 2 / 19

As in the diagram, an inscribed cylinder rests on $z=0$, is coaxial with the paraboloid, and has its top rim on the curved boundary. ThusAllowing contact with the boundary is the usual meaning of inscribed. Form the Lagrange multiplier functionAt a nondegenerate stationary cylinder, its two derivative equations areEliminating $\lambda$ gives $h=H{r}^2/R^2$. The constraint then gives $h=H/2$ and $r=R/\sqrt{2}$, soThe degenerate endpoints $r=0$ or $h=0$ have zero volume, consistent with the positive stationary value being the maximum. If the strict inequalities defining the open interior were interpreted as forbidding boundary contact, the same value would instead be a supremum approached by smaller cylinders.

Use the diagram's labels: $a,b$ are horizontal side lengths and $c$ is the height of the rectangular parallelepiped. Its four top vertices are at horizontal distance $\sqrt{a^2+b^2}/2$ from the axis. Thus the contact constraint and volume areApply a Lagrange multiplier:For positive dimensions, the derivative equations areSubstituting the last equation into the first two givesHence $a=b$, and the constraint becomes $2c=H$. ThereforeThese are the maximal cylinder and box in an elliptic paraboloid dimensions for the rectangular case. A horizontal rotation of the square footprint gives the same volume by rotational symmetry.

The ratio of the two maximal volumes isThus the cylinder has the larger maximum volume, by a factor $\pi /2$. Both optimal shapes have height $H/2$, while their horizontal cross-sections have areas $\pi R^2/2$ and $R^2$, respectively.