Calculate maximum and minimum when second partial derivative test fail 
Calculate the maximum and minimum of function $z = f(x,y) = x^2 - y^2 +2$ subjected to the inequality constraint $D=\{(x,y)| x^2 + \frac{y^2}{4} \leq 1\}$.

My solution:
First form the function
$$
g(x,y) = x^2 + \frac{y^2}{4} - c, ~0 \leq c \leq 1.
$$
Then form the Lagrangian function
$$
L(x,y,\lambda) = x^2 - y^2 + 2 + \lambda\left(x^2 + \frac{y^2}{4} - c\right).
$$
Therefore we have
$$
\left \{
\begin{array}{ll}
L_x' = 2x +2\lambda x = 0\\
L_y' = -2y + \frac{\lambda}{2}y = 0 \\
L_\lambda'  = x^2 + \frac{y^2}{4} - c
\end{array} \right.
$$
After solving above equations, so we can got saddle point like $(\varphi(c),\psi(c))$.

*

*when $c = 0$, we have $x = y = 0$.

*when $c \neq 0$,  there are two kind of solutions:  (1 $x = 0$, we have $y = \pm 2\sqrt{c}$; (2 $y=0$, we have $ x = \pm \sqrt{c}$.

My problem is that i cant use second partial derivative test for judging $(0,\pm 2\sqrt{c})$ and $(\pm \sqrt{c},0)$ are maximum or minimum, obviously $AC-B^2 = 0$. How can i do next? Thanks in advance!

HINT: It is an absolute maximum (and minimum). It is easy to see that it is obtained on the boundary. Then you will have a finite number of possible points. Simply compare them.
