= Solution
Now $d=1/e$, and both constraints are active. Therefore
$$
x_2=\frac1{e^2},
\qquad
x_1=c-x_2=\frac2{e^2}.
$$
The first stationarity equation gives
$$
\lambda=1-\log2>0,
$$
and the second gives
$$
\mu=\frac{2\log2}{e}>0.
$$
Thus all multiplier and complementary-slackness conditions hold, and
$$
\boxed{x_1=\frac2{e^2},
\qquad
x_2=\frac1{e^2}},
$$
with minimum value
$$
\boxed{\frac{2\log2-5}{e^2}}.
$$
The observation is that lowering $d$ activates the second constraint and moves the optimum to the intersection of the two active boundaries. Rewriting $\sqrt{x_2}\leq d$ as $x_2\leq d^2$ 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.
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