Using Lagrange multipliers to find extrema I'm given $f=x^2+4y^2$ defined on disc $x^2+y^2\leq4$
How would I apply Lagrange multipliers to find the extrema of this function?
 A: first of all, you need to find all critical points that lies inside the disk $x^{2}+y^{2}\leq 4$, all you need to is set $f_{x}=f_{y}=0$ and in the worst case scenario you might obtain a system of two equations but here its clear that
$$
f_{x}=0 \implies 2x=0\qquad\text{and}\qquad f_{y}=0\implies 8y=0
$$
Thus, the only critical point we have is the point $(0,0)$. Next, we proceed with Lagrange Multipliers and we treat the constraint as an equality instead of the inequality.(We only need to deal with the inequality when finding the critical points). Applying lagrange multipliers gives us :
$$
f_{x}=\lambda g_{x}\qquad\text{and}\qquad f_{y}=\lambda g_{y}
$$
We also have the fact that $x^{2}+y^{2}=4$. When the above equations are evaluated we get :
$$
2x=2\lambda x \qquad\text{and}\qquad 8y=2\lambda y
$$
Which means :
$$
2x(1-\lambda) = 0 \qquad\text{and}\qquad 2y(4-\lambda)=0
$$
Which means for the first one you have either $x=0$ or $\lambda = 1$ and for the second one you have either $y=0$ or $\lambda = 4$.
For the first one if we substitute $x=0$ in the equality, we will get $0+y^{2}=4 \implies y=\pm 2$. Therefore, we aquired two points $(0,-2)$ and $(0,+2)$. Moreover, if we substitute $\lambda_{1}=1$ in the second equation(not the first), we will get
$$ 2y(4-1)=0\implies 6y=0\implies y=0 $$
and by substituting $y=0$ in the equality, we will get $x=\pm 2$ which means we obtain two new points $(-2,0)$ and $(2,0)$. Hence, we have obtained in total $5$ points that are :
$$
(0,0)\qquad(-2,0)\qquad(2,0)\qquad(0,-2)\qquad(0,2)
$$
All you have to do now is substitute these points in $f$ and not $g$. Now you may be asking well what about $\lambda_{2}=4$ and $y=0$ why didn't we use them? Well if you do use them you will obtain the same critical points we found for $\lambda_{1}=1$ and $x=0$
