Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2019 / Paper 2 / 19

For a closed circular cylinder, $S=2\pi r^2+2\pi rh$ and the volume constraint is $\pi r^{2}h=V$. Use a Lagrange multiplier:The stationary equations areFor $r>0$, the second gives $\lambda =2/r$; the first then gives $h=2r$. The constraint yieldsandTo verify a global minimum, eliminate $h$ to obtain $S(r)=2\pi r^2+2V/r$. It tends to infinity at both ends of $r>0$ and has $S^{\prime \prime }(r)=4\pi +4V/r^3>0$, so its unique stationary point is the global minimum.

The base lies in the plane $z=R-h$. Its circular rim lies on the sphere, soMaximize the cone volume $V_c=\pi r^{2}h/3$ using the Lagrange multiplier expressionFor a nondegenerate cone, the $r$ equation gives $\lambda =\pi h/3$. The $h$ equation is $\pi r^2/3+2\lambda (R-h)=0$, hence $r^2=2h(h-R)$. Combining this with $r^2=2Rh-h^2$ givesIndeed, eliminating $r$ gives $V_c(h)=\pi h^2(2R-h)/3$, which vanishes at the endpoints and has a unique positive interior maximum at $4R/3$. Thus

Because all variables are positive, the product constraint is equivalent to ${\sum }_i\ln x_i=\ln a$. The Lagrange multiplier equations forare $1-\lambda /x_i=0$. Thus every $x_i$ equals $\lambda$, and the constraint requires $\lambda =a^{1/n}$. The candidate minimum is thereforeFor a global justification, put $y_i=\ln x_i$. The objective becomes the strictly convex function ${\sum }_i{e}^{y_i}$ on the affine constraint ${\sum }_i{y}_i=\ln a$. A stationary point of this restricted convex function is its unique global minimum, establishing the claim without presupposing the desired inequality.

Finally take $a={\prod }_i{x}_i$ and divide the minimum bound by $n$. This is the arithmetic-geometric mean inequality:with equality exactly when all the positive variables are equal.