# Find the max and min values for $f(x,y)=y^2-x^2$ subject to $(1/4)x^2+y^2=1$ using the Lagrange Multiplier

lagrange multipliermaxima-minimamultivariable-calculusreal-analysis

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?

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 $$(0, \pm 1)$$.