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For the rotational elliptic paraboloid $z=H(1-(x^2+y^2)/R^2)$ above $z=0$, allow inscribed shapes to touch the boundary. A coaxial cylinder of radius $r$ and height $h$ obeys $h+H{r}^2/R^2=H$. Applying a Lagrange multiplier to $\pi r^{2}h$ gives $h=H{r}^2/R^2$, hence $h=H/2$ and $r=R/\sqrt{2}$.

A centred rectangular parallelepiped with horizontal side lengths $a,b$ and height $c$ obeys $c+H(a^2+b^2)/(4R^2)=H$. The Lagrange multiplier equations for $abc$ give $a=b$ and $c=H{a}^2/(2R^2)$, hence $a=b=R,c=H/2$. These yield the displayed maximal volumes; both positive stationary shapes are interior optima among the dimension variables, while their degenerate boundary shapes have zero volume. The cylinder's volume is $\pi /2$ times the box's. If strict containment in the open region is required, these are suprema approached by slightly smaller shapes rather than attained maxima.

For a closed hemisphere of radius $a>0$, take a rectangular parallelepiped with a face on the flat base, base half-widths $x,y>0$ and height $z>0$. Its curved-boundary constraint is $x^2+y^2+z^2=a^2$ and its volume is $4xyz$. A Lagrange multiplier gives $x=y=z=a/\sqrt{3}$, yielding the displayed maximum. The full edge lengths are $2a/\sqrt{3},2a/\sqrt{3},a/\sqrt{3}$.

This bound remains valid when arbitrary orientations are allowed. Write a cuboid's half-edge lengths as $u_1,u_2,u_3$, its center components in its orthonormal edge basis as $v_i$, and the vertical unit-vector components as $n_i$. Containment above the base plane requires $\sum n_i{v}_i\ge \sum |n_i|u_i$. Hence $|v_j|\ge u_j$ for at least one $j$: otherwise every contributing term would be strictly smaller than its required bound. Its farthest vertex has squared distanceContainment in the sphere bounds the right side by $a^2$. The arithmetic-geometric mean inequality then gives $2u_1{u}_2{u}_3\le a^3/(3\sqrt{3})$, so the full volume $8u_1{u}_2{u}_3$ cannot exceed the displayed value. The constructed cuboid attains it. With strict containment and no boundary contact, the same value is a supremum.

For a lidless rectangular parallelepiped with positive dimensions $l,w,h$ and prescribed external surface area $S=lw+2lh+2wh$, the largest volume is attained at $l=w=2h=\sqrt{S/3}$. The arithmetic-geometric mean inequality applied to $lw,2lh,2wh$ giveswith equality exactly when these three face-area terms are equal. This proves both the claimed global bound and its attainable equality case. Lagrange multipliers produce the same dimensions from the interior stationarity equations. The weighted open-box minimization is the corresponding fixed-volume formulation with general positive face weights.

Take a face on the flat base and let $x,y$ be its positive half-widths, with height $z>0$. The upper vertices meet the spherical boundary at $x^2+y^2+z^2=a^2$, and the rectangular parallelepiped has volume $V=4xyz$. A Lagrange multiplier givesMultiplication by $x,y,z$ respectively shows $x^2=y^2=z^2$. Positivity and the constraint give $x=y=z=a/\sqrt{3}$. ThereforeBoundary choices with a zero dimension give zero volume, so this is the maximum in this configuration.

The maximum-volume cuboid in a hemisphere also has a bound covering arbitrary orientations. Let $u_i$ be any box's half-edge lengths, $v_i$ its centre coordinates in its orthonormal edge basis, and $n_i$ the vertical unit-vector coordinates. Remaining above the base plane requires $\sum n_i{v}_i\ge \sum |n_i|u_i$, so $|v_j|\ge u_j$ for at least one index. Its farthest vertex obeyswhere containment supplies the first inequality. The arithmetic-geometric mean inequality applied to the three terms $u_i^2$ for $i\ne j$ and $4u_j^2$ yields $2u_1{u}_2{u}_3\le a^3/(3\sqrt{3})$. Thus its full volume $8u_1{u}_2{u}_3$ cannot exceed the value attained above; tilting cannot improve it. Boundary contact is allowed for the maximum; strict containment would give the same supremum.

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.