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Consider the function:

$$f(x,y) = x^3 + y^3 + 3xy,\ x,y \in \mathbb{R}$$

Setting the gradient to zero we find that there are two candidates: $\alpha = (-1,-1)$ and $\beta = (0,0)$, looking at the hessian matrix we find that $\alpha$ is a maximum and $\beta $ a saddle point.

but $\alpha$ is actually not only a local but an absolute maximum, how do we determine that in general and specifically in this case?

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How would it be the global maximum? Say, $\lim_{x\rightarrow\infty}f(x,1)=\lim_{x\rightarrow\infty}(x^{3}+3x+1)=\infty$. So for any $M>f(\alpha)=f(-1,-1)$, there is an $(x_{M},1)$ such that $f(x_{M},1)>M>f(\alpha)$, so $f(\alpha)$ is not a global maximum.

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I don't think it is an "absolute" maximum value since for $$x,y\to +\infty \implies f(x,y) \to +\infty$$

To deal with this kind of issue you should refer to the Extreme value theorem .

Notably the existence of a global maximum and a minimum is guaranted if the domain of f is compact (i.e. closed and bounded) and the function is continuos. Otherwise the existence is not guaranted.

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