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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.