Find the critical points of each of the functions below and classify them as a local maximum, minimum, or neither

Question

$f ( x , y ) = \ln ( 2 + \sin ( x y ) )$. Consider only the critical point $(0,0)$.

I have solved for the first and second partial derivatives and I see that they are both equal to $0$ at $(0,0)$. One of the exercises in my textbook mentions the Hessian matrix and I think I should be using that, but I am not sure how it works.

$f ( x , y ) = \left( x ^ { 2 } + 3 y ^ { 2 } \right) e ^ { 1 - x ^ { 2 } - y ^ { 2 } }$

For this problem, I took the first and second partial derivatives and I observed that the critical points are at $(0,1)$, $(0,-1)$, and $(0,0)$. Do I have to use the Hessian here as well? How would that work?

Answer 1

For the first one we have

$$f ( x , y ) -f(0,0)= \ln ( 2 + \sin ( x y ) )-\ln 2=\frac{xy}2+o(x^2+y^2)$$

and then it is a saddle point.

For the second one we have

$$f ( x , y ) = \left( x ^ { 2 } + 3 y ^ { 2 } \right) e ^ { 1 - x ^ { 2 } - y ^ { 2 } }\ge \left( x ^ { 2 } + y ^ { 2 } \right) e ^ { 1 - x ^ { 2 } - y ^ { 2 } }=r^2e^{1-r^2}=er^2-r^2+o(r^2)$$

that is positive as $r\to 0$ and therefore it is a minimum.

Answer 2

Sometimes a graph is worth a thousand words:

As @gimusi points out, this is clearly a saddle point.

Here's your second function:

Clearly a minimum because for any $r>0$, a point a distance $r$ from $(0,0)$ is above the point at the origin. The second derivative with respect to $x$ evaluated at $(0,0)$ is $e$ ($>0$) and hence the origin is a minimum.