Using Lagrange multipliers to find the max and min I have the equation $$f(x,y)=x^2+y^2$$ and the constraint $$(x-1)^2+4y^2=4$$
So I must find the min and max. My try:
So I get the equations:
$$2x=λ(2x-2)$$
$$2y=λ8y$$
and I get λ=1/4 and x=1/3 
Any help is appreciated
 A: Outline
You have set it up right. 
$L = x^2 + y^2 - \lambda\{(x-1)^2 + 4y^2 - 4\} = 0 $ and then find partial derivatives of $L$ with respect to $x$ and $y$. You have made a slight mistake in solving, you should get $\lambda = \frac14$ and $x = -\frac13$. 
Now put back this value of $x$ in the constraint and solve for $y$, you should get $y = \pm \frac{\sqrt5}{3}$. I think you can take it from here.
You may also observe that this only gives the $\color{blue}{minimum}$ value of $x^2 + y^2 = \frac23$.  
To get the maximum value, I would reason out that the constraint is an $\color{blue}{ellipse}$ and we are trying to find the largest $\color{blue}{circle}$ centered at the origin and at least touching, if not intersecting the ellipse. Touching because we want an extrimity. So, this ellipse intersects $X-$axis at $x = -1$ and $x = 3$. So the circle radius will be 3, i.e.. $x = 3$, and in order to satisfy the contraint, $y = 0$. So, the $\color{blue}{maximum}$value is $9$.
Summary
The minimum value of $x^2 + y^2$ is $\color{blue}{\frac23}$ and the maximum value is $\color{blue}{9}$.
A: Actually, there's no need to use multipliers.  You may solve the constraint easily for $y$ and plug this result in $f$. Having done this you'll see nearly instantly the extrema at $x=-1/3,-1,3$, even without using differentiation.
A: Easier to remember general derivation how LM comes about and then apply in our particular case.
$$ f = x^2 + y^2 ; g   = (x-1)^2 + 4y^2 - 4  $$
A LM can be chosen using partial derivatives:
$$ \lambda = \dfrac{ f_x}{ f_y} =   \dfrac{ g_x}{ g_y}. $$
