= Solution
Let $U=\{x\geq0:\|x\|_2=1\}$. This set is nonempty and <compact> for $n\geq1$, and the <quadratic form> is continuous. Its minimum $\alpha$ is therefore attained.
For every nonzero $y$ in the <nonnegative orthant>, normalizing $y$ gives
$$
y^T(Q-\lambda I)y
=\|y\|_2^2\left[\left(\frac y{\|y\|_2}\right)^T
Q\left(\frac y{\|y\|_2}\right)-\lambda\right].
$$
The zero <vector> imposes no further condition. It follows that $Q-\lambda I$ is a <copositive matrix> exactly when $\lambda\leq\alpha$. Thus the feasible <scalar> set is $(-\infty,\alpha]$, and
$$
\boxed{\alpha=\max\{\lambda:Q-\lambda I\in K\}}.
$$
Both optima are attained. This <copositive reformulation of an orthant Rayleigh minimum> follows directly from homogeneity; it does not require invoking a duality theorem for the original nonconvex constraint.
The restriction to $x\geq0$ matters. For the <matrix> in part (a), this minimum is $1$ whereas the unrestricted smallest <eigenvalue> is $-1$.
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