Using Lagrange multipliers to identify the Extremes of function $f(x, y)=x-y$, under condition $g(x,y)=x^2 + y^2 - 4=0$ I'm studying in preparation for a Mathematical Analysis II examination and I'm solving past exam exercises.
If it's any indicator of difficulty, the exercise is Exercise 3 of 4, part $b$ and graded for 10%.

With the Lagrange multiplier method, study the extremes of function $f(x,y)=x-y$ under condition $g(x,y)=x^2 + y^2 - 4 = 0$

I began with noting that $f$ and $g$ are $C^1$ because $f_x= 1,\ f_y= -1,\ g_x = 2x,\ g_y= 2y$ are continuous.
Afterwards, we have: $\nabla f(x,y) = \lambda \nabla g(x,y)$, which, with the help of the condition $x^2 + y^2 - 4 = 0$, leads us to extreme points $(\sqrt{2},-\sqrt{2})$ and $(-\sqrt{2},\sqrt{2})$.
Now, how can I study what these extremes are? Are they local maximums, local minimums? I know that I'm pretty close and I should use determinants $\Delta_1$ and $\Delta_2$, but for what function, $f$ or $g$? And since $f_{xx} = f_{yy} = 0$, would that lead anywhere?
Any help would be greatly appreciated, thank you.
 A: For anyone interested, I was able to come to a seemingly valid solution using Evgeny's help. The reference I used is here. If you find any mistakes in my process, please point it out or write a correct answer, and I'll accept it instead of my own.
Using the Bordered Hessian matrix: $H = \left[\begin{array}{c}0&g_x&g_y\\g_x&f_{xx}+\lambda g_{xx}&f_{xy}+\lambda g_{xy}
\\g_y & f_{yx} + \lambda g_{yx} & f_{yy}+\lambda g_{yy}\end{array}\right]$
In our exercise, it is written as: $H = \left[\begin{array}{c}0&2x&2y
\\2x&2\lambda & 0
\\2y&0& 2\lambda\end{array}\right]$
We calculate $3\times 3$ ${\rm det}H$ as per the reference document, so it goes: ${\rm det}H = 0+0+0 -2y2\lambda 2y - 2x2x2\lambda$. We know from before, when we found the extreme points, that $\lambda = \dfrac{1}{2x} = \dfrac{-1}{2y}$. Using that, we end up with:
${\rm det}H=4y - 4x$
Finally we calculate ${\rm det}H(\sqrt{2},-\sqrt{2}) = -8\sqrt{2}$ and ${\rm det}H(-\sqrt{2},\sqrt{2}) = +8\sqrt{2}$. As the PDF says, since the determinant of the Hessian matrix is negative in the Lagrange critical point $(\sqrt{2},-\sqrt{2})$ is negative, that extreme is minimum. With the same process, $(-\sqrt{2},\sqrt{2})$ is a maximum because the determinant of the Hessian matrix is positive.
A: In my opinion, you don't need to use the Bordered Hessian for this question. Your objective function $f(x,y)=x-y$ describes a plane passing orijin and your constraint $x^2 + y^2 =2^2$ is a circle with radius $2$ and center at $(0,0)$. Hence geometrically you have one maximum and one minimum point being symmteric with respect to orijin. You found them as $(\sqrt{2}, -\sqrt{2})$ and $(-\sqrt{2}, \sqrt{2})$ by Lagrange method. Now just put them into the function and compare them to which one is maximum and which one is minimum:
$f(\sqrt{2}, -\sqrt{2}) = \sqrt{2} - (-\sqrt{2}) = 2 \sqrt{2}$
$f(-\sqrt{2}, \sqrt{2}) = -\sqrt{2} - (\sqrt{2}) = - 2 \sqrt{2}$
Thus, $(\sqrt{2}, -\sqrt{2})$ is the maximum point and $(-\sqrt{2}, \sqrt{2})$ is the minimum point.
Please note that here the geometry matters. Your constraint is a circle which is closed and bounded. Also, $g(x,y) \leq 0$ is convex. These simplify many things in general and make geometric interpretations clear. Otherwise, you need some other tools like Bordered Hessian as @Evgeny suggested.
