I am working on a Lagrange multipliers problem where I have to find the extreme values of the function $f(x, y) = x^2 + y^2$ along the constraint curve $g(x, y) = x^4 + y^4 = 1$.
Solving the Lagrange equations I get $\lambda = \frac{1}{2x^2}$ and $\lambda = \frac{1}{2y^2}$ i.e. $x = \pm y$. However, when plugging this result back into the constraint curve and solving for $g(x, y)$, I only get one extreme value, $\sqrt{2}$.
Solutions for this problem say that when you obtain the Lagrange equations you can also assume that $x$ or $y$ are equal to $0$. And when you plug these results back into the constraint curve, you get a minimum of $1$ when one of the variables equals $0$ and the other equals $\pm 1$.
What I am curious about is why exactly you are able to assume the $x$ or $y$ are equal to $0$ in this situation. My textbook says nothing about it and I can't find any examples of a problem where someone explains this. (The solution I found just assumed $x$ or $y$ equaled $0$ without any explanation.) My intuition is that it has something to do with the fact that $x$ and $y$ are in the denominator in this situation, but I'm not sure.
If someone could clear this up for me it would be greatly appreciated.
The problem is that you don't quite end up with
$$\lambda = \frac{1}{2x^2}$$
You end up with
$$2x = 4x^3\lambda$$
You can only divide by $4x^3$ if $x\neq 0$, otherwise the equality is true regardless of $\lambda$. So the possible solutions are
$$\lambda = \frac{1}{2x^2}\ \mathrm{or}\ x=0$$