Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2005 / Paper 2 / 4

The Lagrange multiplier equations are $2x=2\lambda x$, $-2y=2\lambda y$, together with $x^2+y^2=1$. If $x\ne 0$, then $\lambda =1$ and $y=0$; if $y\ne 0$, then $\lambda =-1$ and $x=0$. ThusThe constraint is compact and its gradient never vanishes, so these exhaust the possible global extrema. Equivalently, substituting $x=\cos \theta$, $y=\sin \theta$ gives $z=\cos 2\theta$ along the boundary.

Minimize the squared distance $D^2=(x-x_0{)}^2+(y-y_0{)}^2$ subject to $g=y-mx-c=0$. The Lagrange multiplier condition $\nabla D^2=\lambda \nabla g$ gives $x=x_0-m\lambda /2$ and $y=y_0+\lambda /2$. Enforcing the line constraint gives $\lambda =2(m{x}_0+c-y_0)/(1+m^2)$, henceThe displacement to this point is parallel to the line's normal vector $(-m,1)$, so it is the orthogonal projection. The squared distance restricted to the line has positive quadratic coefficient $1+m^2$, proving this stationary point is the unique global minimum.

Let $w>0$ denote the stated cost per area of the cylindrical wall. Charge for the curved wall and cap as specified; no floor cost is supplied. The volume and cost are $V=\pi r^{2}h+2\pi r^3/3$ and $C=2\pi wrh+4\pi w{r}^2$. To minimize cost at fixed volume, apply a Lagrange multiplier to $C-\lambda (\pi r^{2}h+2\pi r^3/3-V)$. Differentiation with respect to $h$ gives $\lambda =2w/r$. The $r$ derivative then gives $2\pi wh+8\pi wr=\lambda (2\pi rh+2\pi r^2)$, reducing to $h=2r$. Thus the optimal cylinder and hemispherical cap with unequal surface costs hasFor a global check, eliminate $h$ to obtain $C(r)=2wV/r+(8\pi w/3)r^2$. Its second derivative is positive for $r>0$, and its critical radius gives $h>0$, so the constrained optimum lies inside the physically feasible range and is the unique minimum.