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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.


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There is a "second derivative test" that uses the Hesse matrix of a function. – Hans Engler Jan 15 '13 at 23:26

1 Answer 1

up vote 1 down vote accepted

Hint: Evaluate the determinant of the Hessian matrix at your point. (Second Derivative test). I.E.


This 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)



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 $(6,6)$.

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I get it, in terms of the limit however, i dont understand how taking the limit as it goes to infinity will prove its a minimum and not a maximum @rustyn – user5208 Jan 16 '13 at 0:02
@ReubenPereira I have edited/updated my answer. – Rustyn Jan 16 '13 at 3:33

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