Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2013 / Paper 1 / 20

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.

The triangular roof has height $(w/2)\tan \theta$ and two slopes of length $(w/2)\sec \theta$. Hence the area constraint and cost areThe wall height is fixed, so its cost is constant. Assume the physical case $A,h,\beta >0$. Minimize $C$ subject to the area constraint by Lagrange multipliers:Eliminating the positive common factors givesThus the optimum has $0<\theta <\pi /4$ andSet $D=A-wh>0$, so the area constraint gives $\tan \theta =4D/w^2$. Substituting into the preceding equation and multiplying by $w^3$ yields $w^4-16D^2=16hwD$. ThereforeFor a global check, the square of the total roof length isIts derivative has the sign of $P(w)=w^4+16Ahw-16A^2$. Since $P^{\prime }(w)>0$, $P(0)<0$ and $P(A/h)>0$, it has one root in this interval, with $Q$ decreasing before that root and increasing afterwards. This proves the unique global minimum, rather than merely a stationary design. The minimum roof length at fixed wall height and cross-sectional area is independent of $\alpha$ and of the positive factor $\beta$. If $\beta =0$, every feasible shape has equal cost; the displayed conditions then describe a possible minimizer but are not necessary.