Past exam of the mathematics course of the University of Cambridge / 2014 / ib / Paper 1 / 8H / Solution

A useful form of the Lagrangian sufficiency theorem is as follows. For minimization of $f(x)$ subject to $g_j(x)\ge 0$ and $h_i(x)=0$, formIf $x_{\ast }$ is feasible, ${\lambda }_j\ge 0$, ${\lambda }_j{g}_j(x_{\ast })=0$, and $x_{\ast }$ globally minimizes $L$ over the ambient domain for these fixed multipliers, then $x_{\ast }$ is a global constrained minimizer. Indeed, for every feasible $x$,The first inequality uses multiplier signs; the last equality uses feasibility and complementary slackness. Equality constraints alone require no sign restriction on their multipliers. A stationary point of a differentiable convex Lagrangian is a global minimum, which is a common way to verify the theorem's hypothesis.

Here takeStationarity gives $4x_1=4x_2=2x_3=\lambda$. Feasibility then gives $\lambda =1$ and $x_{\ast }=(1/4,1/4,1/2)$. For that multiplier,The Lagrangian has its unique global minimum at $x_{\ast }$, so the sufficiency theorem proves