Hello, I am not sure if I can assume if this function will attain a global extrema or not. I know that if a function is compact and continuous then it will for sure attain a global maxima by the extreme value theorem. However, in this particular example, g has a upper bound but doesn't have a lower bound. Thus, I'm not sure if there will be a global maximum since it only has an upper bound and it is not compact. Does having an upper bound guarantees that the function will attain a maximum on the region g(x,y,z) ? I found that the only critical point satisfying this constraint is (0,0,2). should I assume directly that is a maximum ? or it isnt always the case?
You have inequality constraints which mean that either the extrema occurs on the boundary of these constraints or the interior. If the extrema occurs at the boundary, a dose of lagrange multipliers gives you the solution $(x,y,z,\lambda) = (0,0,2,-4)$. There's a trick for checking whether this point is a maxima or minima. Consider the Lagrange Function $\Lambda = f - (-4) g$ and consider it's hessian with respect to $(x,y,z)$ at the point $(0,0,2)$: $$\textbf{H}_\Lambda = \left[\begin{matrix}10 & 0 & 0\\0 & 10 & 0\\0 & 0 & 2\end{matrix}\right]$$ If this matrix is positive definite (which it is in our case since it's eigenvalues are all positive), then the point we found $(0,0,2)$ is a minimum. Correspondingly, the value of $f$ at this point is $f(0,0,2) = 4$.
If instead the extrema occurs at the interior, then it must be an unconstrained local extrema of $f$ that happens to lie within the interior. Of course, our function $f$ has the property that: $$\nabla f = \left[\begin{matrix}2 x\\2 y\\2 z\end{matrix}\right] = \textbf{0}$$ $\implies x=y=z = 0$ is the only local (and therefore global) extremum of $f$ and it must be a minimum because $f$ is unbounded above. It's also easy to see that $(0,0,0)$ doesn't lie in the set because $x^2 + y^2 - z = 0 \nleq -2$. So that means the global extrema of $f$ on the set (which should be the extreme-most value from the set of extrema we've found on the set) is $f(0,0,2) = 4$ (which happens to be the only extrema we found in the set), and it's characterization is a minimum
Edit: Reviewing the problem, I noticed that we can get around inequality constraints by turning them into equality constraints by considering the same problem in a higher dimension. If $g(x,y,z):= x^2 + y^2 -z \leq -2$, then we can convert this to $g(x,y,z,\alpha):= x^2 + y^2 -z + \alpha^2 = -2$ for some value of $\alpha$. Then we set up the Lagrange function: $\Lambda = f - \lambda g$ and set $\nabla \Lambda = 0$ where $\nabla$ also contains the term for $\frac{\partial}{\partial \alpha}$ since we're in a higher dimensional space now. We get: $$\nabla \Lambda = \left[\begin{matrix}- 2 \lambda x + 2 x\\- 2 \lambda y + 2 y\\\lambda + 2 z\\- \alpha^{2} - x^{2} - y^{2} + z - 2\\- 2 \alpha \lambda\end{matrix}\right] = \textbf{0}$$ Solving this gives us $(x,y,z,\lambda,\alpha) = \left( 0, \ 0, \ 0, \ 0, \ \pm \sqrt{2} i\right), \left( 0, \ 0, \ 2, \ -4, \ 0\right), \left( 0, \ \pm \frac{\sqrt{10} i}{2}, \ - \frac{1}{2}, \ 1, \ 0\right)$. Getting rid of complex solutions leaves us with just: $(x,y,z,\lambda,\alpha) = \left( 0, \ 0, \ 2, \ -4, \ 0\right)$ which means this is the only solution on the boundary of $g(x,y,z,\alpha)=-2$ and therefore the only solution of the problem on the original constraint $g(x,y,z) \leq -2$.