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!
 A: Clearly, $\nabla f(x,y)=0\iff(x,y)=(0,0)$, but $(0,0)$ is a saddle-point. So, the maximum and the minimum can be attained only at the boundary of $D$.
Let $h(x,y)=x^2+\frac{y^2}4$. Then apply the method of Lagrange multipliers:$$\left\{\begin{array}{l}f_x(x,y)=\lambda h_x(x,y)\\ f_y(x,y)=\lambda h_y(x,y)\\h(x,y)=1,\end{array}\right.$$or$$\left\{\begin{array}{l}2x=2\lambda x\\-2y=\frac12\lambda y\\x^2+\frac{y^2}4=1.\end{array}\right.$$It's solutions are $(x,y,\lambda)=(0,-2,-4)$, $(x,y,\lambda)=(0,2,-4)$, $(x,y,\lambda)=(-1,0,1)$ and $(x,y,\lambda)=(1,0,1)$. Since $f(0,\pm2)=-2$, and $f(\pm1,0)=3$, the maximum is $3$ and the minimum is $-2$.
A: Since the the maximum and the minimum can be attained only at the boundary we can proceed by direct substitution of the constraint (ellipse) that is
$$h(y) = 1-\frac{y^2}4 - y^2 +2=3-\frac 54 y^2$$
therefore we have

*

*minimum $f(x,y)=3$ at $y=0$, $x=\pm 1$

*maximum $f(x,y)=-2$ at $y=\pm 2$, $x=0$
A: 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.
