Nonlinear optimization question For (x,y) in $\mathbb R^2$, consider f(x,y) = $x^2 -2xy + \frac{4}{3}y^2 - 4y$ Find the local minimum of f. Is it a strict local minimum? Compute the $\lim\limits_{|(x,y)|\to \infty}$ f(x,y) to decide if the local minimum is a global minimum.
My work: I found the critical point (6,6) and for any v (since it is in the interior of $\mathbb R^2$) is feasible. Therefore $\vec\nabla$F.v = 0 since (6,6) is in $\mathbb R^2$.
I'm having trouble how taking the shown limit will prove whether the local min is a global min.
Thanks,
 A: Hint: Evaluate the determinant of the Hessian matrix at your point. (Second Derivative test). I.E.  
$f_{xx}f_{yy}-f_{xy}^2|_{(6,6)}$ 
This PDF: http://www.math.osu.edu/~kwa.1/254au10/14.7.pdf will give you a nice in-depth explanation of the second derivative test. 
This is for your first question. 
As per your second question: 
$\lim_{|(x,y)|→∞} f(x,y) = \infty$ 
Evaluating the limit we see $\lim_{|(x,y)|→∞} f(x,y)= \lim_{|(x,y)|→∞} x^2 +\frac{4}{3}y^2 - (2xy + 4y)$ ~ $\lim_{|(x,y)|\to \infty} x^2 + \frac{4}{3}y^2 = \infty$ 
We get to do this because $x^2 + \frac{4}{3}y^2 $ > $ (2xy + 4y)$
for all values outside of and on the boundary of the ellipse. (See picture for more info)  

Credits: http://www.wolframalpha.com  
Explicitly: $x^2 + \frac{4}{3}y^2 $ is growing a lot faster than the other term: $2xy + 4y$,  (This does not prove it is an absolute minimum, I just was showing you the limit.)
Explicitly, for all $(x,y)$, we have:
$$
f_{xx}\mid_{(x,y)} = 2, \;\;\; f_{yy}\mid_{(x,y)} = \frac{8}{3},\;\;\; f_{xy}\mid_{(x,y)}= -2
$$
This implies: 
$$
D(H)\mid_{(x,y)} = f_{xx}f_{yy} - f_{xy}^2 = \frac{16}{3} - 4 = \frac{4}{3}
$$
Where $D(H)$ is the determinant of the hessian matrix. Since $D(H)\mid_{(6,6)}>0$ and $f_{xx}\mid_{(6,6)} > 0$ we have a local minimum. 
Moreover, since both $f_{xx}$ and $f_{yy}$ are positive at the critical point and the discriminant is positive, the
surface must be concave upward in every direction.
Thus, this must be an elliptic paraboloid with a global minimum at $(6,6)$. 
