Why the method of Lagrange multipliers fails in this case

Question

I need to minimize and maximize $f(x,y,z)=xy^2z^3$, given that $x^2 + y^2 + z^2 = 6$.

According to the Lagrange multipliers calculator, there is an infinite number of points, where the function achieves the zero value.
But zero is neither minimum, nor maximum: $f(2,1,1)=2$, $f(-2,1,1)=-2$.

Why did the method fail to find the real maximum and minimum, and find the points, that are neither maximum, nor minimum?

Answer 1

It seems the online calculator fails, indeed from the given equation for the case $xyz \neq 0$ we find by elimination of $\lambda$

• $2x^2=y^2$
• $2z^2=3y^2$

that is from the constraint

• $x=\pm 1$
• $y=\pm \sqrt 2$
• $z=\pm \sqrt 3$

To check this result we can use the constraint to obtain the equivalent problem

$$g(x,z)=xz^3(6-x^2-z^2)$$

which indeed has maximum at $(x,z)=(\pm 1,\pm \sqrt 3)$ and minimum at $(x,z)=(\mp 1,\pm \sqrt 3)$.