Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 2 / 7H (Optimisation)

Use multipliers $\lambda ,\mu \ge 0$ forAt an interior point in the nonnegative quadrant, the Karush-Kuhn-Tucker conditions arewith complementary slackness for the two constraints.

For $c=3/e^2$ and $d=2/e$, the square-root constraint is inactive at the answer, so $\mu =0$. The second stationarity equation gives $\lambda =1$, and the first gives $x_1=e^{-2}$. The sum constraint is active, henceIndeed, $\sqrt{x_2}=\sqrt{2}/e<2/e$. The minimum value is

Now $d=1/e$, and both constraints are active. ThereforeThe first stationarity equation givesand the second givesThus all multiplier and complementary-slackness conditions hold, andwith minimum value

The observation is that lowering $d$ activates the second constraint and moves the optimum to the intersection of the two active boundaries. Rewriting $\sqrt{x_2}\le d$ as $x_2\le d^2$ shows that the feasible set is convex, while $x_1\log x_1-x_2$ is convex. Hence the active-set transition in capped resource allocation and the KKT candidates above give the unique global minima, not merely local stationary points.