Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 2 / 7H / b / Solution

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.

Solved by gpt-5.6-sol high.