Why does a variable substitution not preserve boundary conditions? I am playing around with lagrange multipliers and have run into a feature I don't totally understand. I'm thinking about situations where I might just use substitution, and I've found sometimes the solution set with this method misses answers "on the boundary". My question is basically an extension of this stack exchange post. Very simply, consider optimizing $f(x,y)$ on the unit circle, i.e.
$$
f(x,y) = -3x^2 - 10 y^2 \\
g(x,y) = x^2 + y^2 -1.
$$
This problem has solutions at $(0,\pm1)$ and $(\pm1, 0)$. If I substitute $x^2 = 1-y^2$, then I get $\tilde{f}(y) = -7y^2 -3 $, which only yields the $y=0$ solutions (the other two solutions are found if substituting for $x$).
I understand after reading the post above, that the issue is $\tilde{f}$ is not constrained by $y<1$, which is of course true on the unit circle. However, I am interested in why this is the case? Doesn't the function $x^2 = 1 - y^2$ have solutions only on the unit circle (at least in $\mathbb{R}$), why does substituting this condition not preserve "all the information"?

You should be optimizing $\tilde{f}(y)$ with the constraint $x^2 = 1-y^2$. Since $x$ does not appear in the objective function, the constraint simply implies $1-y^2 \ge 0$. So it does contain "all the information"; your mistake was discarding the constraint $x^2 = 1-y^2$ when considering the optimization of $\tilde{f}$.
