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I am working on a problem from my textbook on Lagrange Multipliers. I feel I have these down now, but I am curious about this specific problem. Let \begin{align} f\left(x,y\right)=x^2-y^2\tag{1},\\ \text{and}\;g\left(x,y\right):=2y-x^2=0.\tag{2} \end{align} The first part is easy. Since $g$ is the "constraint" as they call it, we have \begin{align} \vec{\nabla}f\left(x,y\right)&=\begin{bmatrix}2x\\-2y\end{bmatrix},\tag{3}\\\lambda\vec{\nabla}g\left(x,y\right)&=\begin{bmatrix}-2x\lambda\\2\lambda\end{bmatrix}.\tag{4} \end{align} This gives us the set of equations \begin{align} 2y-x^2&=0,\tag{5}\\ 2x&=-2x\lambda,\tag{6}\\ -2y&=2\lambda\implies \lambda=-y,\tag{7} \end{align} and substituting $\left(7\right) $ into $\left(6\right)$ gives us $y=1$, meaning $x=\pm\sqrt{2}$. Then putting these points into the original equation $\left(1\right)$ gives me $\left\{\left(\sqrt{2},1,1\right),\left(-\sqrt{2},1,1\right)\right\}$, and I have graphed to confirm.

But now what about $\left(0,0,0\right)$? It appears to be a relative minimum when I graph these two surfaces, but how do I get it to show up in my Lagrange Multipliers (unless it is not done in that way). I do notice initially that $\left(0,0,0\right)$ does satisfy both equations $\left(1\right)$ and $\left(2\right)$, however. But it does not seem this is impetus for being min alone.

Thank you for your time,

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You have to make sure when solving the equations that you are not dividing by something that is zero. Having done your substitution for $\lambda$, you are left with $$ y-x^2=0 \\ 2x=2xy, $$ and if $x=0$, the second equation is satisfied, and then the first equation gives $y=0$.

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