Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2012 / Paper 1 / 20

Take positive dimensions measured in metres. The volume is $V=lwh$ and the external surface area constraint is $S=lw+2lh+2wh=12$. There is one base and four walls, with no lid. The Lagrange multiplier equations for $V-\lambda (S-12)$ areThe first equation implies $\lambda >0$. Multiply the first by $l$ and the second by $w$ and subtract; this gives $2\lambda h(l-w)=0$, hence $l=w=s$. The last equation gives $\lambda =s/4$. Substituting into the first yields $sh=(s/4)(s+2h)$ and therefore $s=2h$.

The constraint now reads $s^2+4sh=3s^2=12$. ThusFor a global check, apply the arithmetic-geometric mean inequality to $lw$, $2lh$ and $2wh$, whose sum is $12$. Their product is $4V^2\le 4^3=64$, so no feasible box has $V>4$. Equality holds exactly at the dimensions found above. This is the maximum-volume open rectangular box at fixed surface area.

The curve has $x\ne 0$ and can be written $y=1/x-x$. It is a hyperbola with vertical asymptote $x=0$ and oblique asymptote $y=-x$, crossing the horizontal axis at $(\pm 1,0)$. It is symmetric under $(x,y)\mapsto (-x,-y)$.

On each branch $dy/dx=-1/x^2-1<0$. For $x>0$, the branch descends from $+\infty$ as $x\downarrow 0$, passes through $(1,0)$, and approaches $y=-x$ from above as $x\to +\infty$. The negative branch follows by the central symmetry.

Minimize the squared distance $F=x^2+y^2$, which has the same minimizers as the distance. The Lagrange multiplier equations for the constraint $G=x^2+xy-1=0$ areThe constraint gradient cannot vanish on this branch, since its second component is $x>0$. Eliminate $\lambda =2y/x$ to obtain $x^2=2xy+y^2$. Put $t=y/x$; then $t^2+2t-1=0$, so $t=-1\pm \sqrt{2}$. But the constraint says $x^2(1+t)=1$, requiring $1+t>0$ and selecting $t=\sqrt{2}-1$. It follows thatTo prove a global minimum on the whole $x>0$ branch, substitute $y=1/x-x$ into $F$:The arithmetic-geometric mean inequality proves this bound, with equality only when $2x^2=x^{-2}$, namely $x=2^{-1/4}$. Thus the stationary point is the unique closest point on the required branch. Numerically it is approximately $(0.840896,0.348311)$ and its distance is $0.910180$ in coordinate units.