I need to minimise $$\begin{split}f(x,y)=2x+xy+3y\\ s.t.\begin{cases} x^2+y\geq3\\ x+0.5\geq 0\\ y\geq0\end{cases}\end{split}$$ I get my lagrangian, $$L(x,y,\lambda,\mu) = -\lambda x^2 - \lambda y + 3\lambda - 0.5 \mu - \mu x + 2x + xy + 3y.$$
I use the Kuhn Tucker conditions for finding a min: $$\begin{array}{c c c} L_x\geq 0 & x\geq 0 & xL_x=0\\ L_y\geq 0 & y\geq 0 & yL_y=0\\ L_{\lambda}\leq 0 & \lambda\geq 0 & \lambda L_{lambda}=0\\ L_{mu}\leq 0 & \mu\geq 0 & \mu L_{\mu}=0 \end{array}$$ I get 3 points: \begin{array}{|r|r|r|r|} \hline x & y & \lambda & \mu \\ \hline 0 & 3 & 3 & 0 \\ 0.333333 & 2.88889 & 3 & 0 \\ 1.732051 & 0 & 0.577350 & 0\\ \hline \end{array}
My question is, the constraint $x+0.5\geq 0$ is non-binding because of the KT conditions require $x$ in the positive orthant, but surely I can not just completely ignore this constraint. Further, the latter of my $3$ answers yields the lowest value of $f(x,y)$. Is this my minimum? How do I distinguish between the points?
Many thanks
As $y\geq 0$, you have that $x\geq \sqrt{3}$, therefore, the constraint $x+0.5\geq 0$ will never be binding and can be safely ignored. In other words, $y\geq 0\wedge x^2+y\geq 3\implies x-\frac{1}{2}\geq 0.$
In order to know which of the points are local minima, you can check the second order conditions. If you're interested in the global minimum, you need, as you did, check where does the function attain its lowest value.