Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 38 / 1

For a maximization problem with $g_i(u)\le 0$ and $h_j(u)=0$, use the optimization LagrangianThe Lagrangian sufficiency theorem says: if $u^{\ast }$ is feasible, $u^{\ast }$ globally maximizes $L(\cdot ,{\lambda }^{\ast },{\mu }^{\ast })$ over its original domain, and complementary slackness holds, ${\lambda }_i^{\ast }{g}_i(u^{\ast })=0$, then $u^{\ast }$ globally maximizes $f$ over the feasible set. Equality Lagrange multipliers have no sign restriction. For every feasible $u$,which proves the theorem.

For a minimization problem, reverse the signs in the optimization Lagrangian: take $L=f+{\sum }_i{\lambda }_i{g}_i+{\sum }_j{\mu }_j{h}_j$, with ${\lambda }_i\ge 0$. If a feasible $u^{\ast }$ globally minimizes this optimization Lagrangian and satisfies complementary slackness, thenThe hypothesis is a global extremum of the Lagrangian. Merely solving its stationarity equations is insufficient; no convexity assumption is needed when the global extremum itself has been proved.

With $g=2x-y-z-2$ and $h=x^2+y^2-5$, the maximization optimization Lagrangian isA finite unconstrained maximum requires $\lambda =1$ to cancel the coefficient of $z$. With $\mu =1$, completing the square givesThus its global maximizers have $x=-1$ and $y=2$. The equality constraint holds, and complementary slackness with $\lambda =1$ makes $g=0$, giving $z=-6$. The Lagrangian sufficiency theorem certifiesEvery feasible point has objective at most $L\le 12$, so this is a global conclusion rather than just a stationary-point calculation.

For minimization, the optimization Lagrangian isIts coefficient of $z$ is $-1-\lambda <0$, so its infimum over unrestricted $z$ is $-\infty$ for every allowable Lagrange multiplier. There is no finite global minimizer of the optimization Lagrangian to which the Lagrangian sufficiency theorem could apply.

The primal minimization problem is also unbounded: set $x=0$, $y=\sqrt{5}$ and $z=t\ge 0$. These points are feasible, and $3y-z=3\sqrt{5}-t\to -\infty$. Hence

For any feasible $x,y$, increasing $z$ lowers the objective, so the additional upper bound makes $z^{\ast }=0$. Feasibility then requires $2x-y\le 2$. The lowest feasible value of $y$ on the circle occurs at the lower intersection with $2x-y=2$. Solving gives the candidateA global certificate avoids relying on the circle sketch. Introduce Lagrange multipliers for $g\le 0$, $z\le 0$ and $h=0$:ChooseThe coefficient of $z$ vanishes, and $2\mu x^{\ast }+2\lambda =0$, $2\mu y^{\ast }+3-\lambda =0$. Therefore completing the square yieldsThe candidate minimizes $L$ globally and satisfies both active inequality constraints, so complementary slackness and the Lagrangian sufficiency theorem give