Min$_{x}~x$
Subject to $x \geq 0$
For this problem, is $(x^{*}, \lambda^{*})=$$(0,0)$ a KKT point ?
My try : I formulated corresponding Lagrangian and tried to find out the KKT point(s). But it is not giving $(0,0)$. But if we notice, $x=0$ is the solution for for the given constrained optimization problem.
No. The KKT point is $(x^*,\lambda^*)=(0,1)$. $\lambda=0$ is not dual feasible.
The Lagrangian is $L(x,\lambda)=x-\lambda x$, and the dual problem is $$\begin{array}{ll} \text{maximize} & 0 \\ \text{subject to} & \lambda = 1 \\ & \lambda \geq 0 \end{array}$$ So clearly, $\lambda^*=1$ is the optimal dual point.
It's actually not difficult to see why this is the case if you consider the dual cost interpretation. If the primal model is changed to $x\geq \epsilon$, where $\epsilon$ is small, then the objective should change by $\lambda\epsilon$. Since this perturbation yields $x^*=\epsilon$, it is clear that $\lambda^*=1$.