Past exam of the mathematics course of the University of Cambridge / 2020 / ib / Paper 2 / 19H / Solution

The Lagrangian sufficiency theorem for a maximization problem says the following. Suppose the constraints are $g_i(q)\ge 0$ and $h_j(q)=0$, and defineIf a feasible $q^{\ast }$ satisfies complementary slackness ${\lambda }_i{g}_i(q^{\ast })=0$ and globally maximizes $L(\cdot ,\lambda ,\mu )$, then $q^{\ast }$ globally maximizes $f$. Indeed, for every feasible $q$,This proves the theorem. In a concave maximization problem, stationarity and the boundary optimality conditions ensure the required global maximum of the Lagrangian.

For the problem at hand, retain $x,z\ge 0$ as the domain and attach a multiplier $\lambda$ to the equality:For $\lambda \ge 1$, this is concave in $(x,y,z)$ on the domain. Its separate maximizers satisfywhile the $x$ term is $(1-\lambda )x$. Thus $x$ may be positive only when $\lambda =1$; when $\lambda >1$, its maximizing value is $x=0$.

Iftake $\lambda =1$. Then $y=1$, $z=a^2-1$, and feasibility fixesThe Lagrangian sufficiency theorem gives the maximum

Ifthen $\lambda >1$ and $x=0$. The equality constraint becomessoConsequentlyThe assumptions $a\ge 1$ and $b\ge 1/2$ ensure $z\ge 0$. At the stationary point $\sqrt{1+z}=ay$, so the maximum value isThe two formulas agree at $b=a^2-1/2$.