Use multiplier $-z$ for $Ax-b-s=0$. The Lagrangian is
$$
L(x,s,z)=\frac12x^TQx+I_{\mathbb R_+^m}(s)-z^T(Ax-b-s).
$$
The infimum over $s$ is finite exactly when $z\geq0$. The infimum over $x$ occurs at $x=Q^{-1}A^Tz$, and hence
$$
\boxed{h(z)=b^Tz-\frac12z^TAQ^{-1}A^Tz.}
$$
The dual is $\max_{z\geq0}h(z)$. Since the primal objective is coercive, the explicit <Slater condition>
$$
\boxed{\text{there exists }\bar x\text{ such that }A\bar x>b}
$$
is sufficient for feasibility, attainment, and equality of primal and dual values.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 339 / 2 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-339/2/b/solution