while solving some exercises related to finding maximums and minimums of functions with two variables under constraints, I encountered a doubt when using Lagrange multipliers.
For instance, considering the function ( f(x,y,z) = xyz restricted at x^2 + 2y^2 + 3z^2 = 6 ), I found the following system:
\begin{align*} yz &= \lambda 2x \\ xz &= \lambda 4y \\ xy &= \lambda 6z \\ \end{align*}
Subsequently, I obtained x = sqrt(2), y = + or - 1, and z = + or - sqrt(2/3).
I know that if I take the point (sqrt(2), 1, sqrt(2/3)), which is given in the answer key, it satisfies the constraint equation and seems to be a maximum point (value 2/sqrt(3)). However, this point, when substituted into xy = λ6z, does not satisfy the equality.
Hence, my question is: shouldn't a point that indeed represents an extreme under the function and constraint conditions satisfy both equations?