When you solve the system of the derivates, you end up with two different values for the same multiplier as well as $-1/2 = 2$
This is my solution:
$\nabla f(x,y)=\lambda \nabla g(x,y)$ which takes us to the following nonsense system:
$-2x=\lambda (1/2)x$
$2y=\lambda2y$
it gives us $\lambda = 4$ and $\lambda = 1$ and $-0,5=2$
How can this system of equations give me the maximum and minimum points that I'm looking for?
Best Answer
Apart from solving the system that gets the stationary points of
$$L(x,y,\lambda) = y^2-x^2- \lambda\left(\frac 14 x^2 +y^2-1\right),$$
you must make the argument that $f$ actually attains a max/min on the admissible region, and that these max/min are critical points of the Lagrangian function.
The existence of max/min comes from the fact that $f$ is continuous over the admissible region and that the admissible region (defined by $\frac 14 x^2+y^2=1$) is a compact set.
The fact that all involved functions are differentiable and that the Jacobian matrix of the restriction has maximal rank allows you to conclude that the min/max are attained at critical points of $L$.
Just solving the system is not a complete answer.
Regarding the solution of the system, you have that $$ -2x-\lambda \frac x2=0, \quad 2y -2\lambda y = 0,\quad \frac 14 x^2+y^2=1$$
Starting with the first equation, you get $x=0 \vee \lambda = -4$. If $x = 0$, you get $y = \pm 1$. If $\lambda = -4$, you get $y=0$ and then $x = \pm 2$. So, four candidates for min/max: $(0, \pm 1)$ and $(\pm 2, 0)$. Checking the value of the objective function $f$ over these four points, you conclude that the global minimum is attained at $(\pm 2 0)$ and the global maximum is attained at $(0, \pm 1)$.