 Quadratic minimization with a moving upper bound

Consider the minimization problemThe objective function is a strictly convex function with its unconstrained optimization minimum at $x=2$. If $b<2$, it decreases throughout the feasible set until the bound $x=b$ is reached. If $b\ge 2$, the feasible set already contains $2$. HenceWith $g(x)=x-b$, the optimization Lagrangian and KKT conditions areSubstituting $x_{\ast }(b)$ gives the unique inequality Lagrange multiplierFor $b<2$, the active constraint has ${\lambda }_{\ast }>0$. For $b>2$, there is strict slack $b-x_{\ast }=b-2$ and ${\lambda }_{\ast }=0$. At $b=2$, the active constraint has zero Lagrange multiplier: the boundary happens to pass through the unconstrained optimum. Feasibility and complementary slackness must both be checked; solving only $2(x-2)+\lambda =0$ would allow infeasible candidates.

An independent certificate follows by completing the square:The Lagrange dual function is a concave function of $\lambda$. Its maximum over $\lambda \ge 0$ occurs at ${\lambda }_{\ast }(b)$ and equals $V(b)$. Thus every case has an explicit lower bound attained by a feasible point; the KKT conditions provide a certificate that can be checked without searching all real numbers.