Finding both max and min points by Lagrangian Using the method of the Lagrange multipliers, ﬁnd the maximum and minimum values of the function
$$f(x,y,z) = x^2y^2z^2$$
where $(x,y,z)$ is on the sphere
$$x^2 +y^2 +z^2 = r^2$$
So $L(x,y,z;\lambda)=x^2y^2z^2 + \lambda (x^2 +y^2 +z^2-r^2)$. 
After finding $L_{x,y,z,\lambda}=0$, I got $$\bigg(\pm \frac r {\sqrt 3},\pm \frac r {\sqrt 3},\pm \frac r {\sqrt 3}\bigg)$$ which seems like the maximum but what is the minimum?
 A: Let $g(x,y,z) = x^2 + y^2 + z^2$.
Writing $\nabla f = \lambda \nabla g$ leads to
$$
xy^2z^2 = \lambda x \\
x^2yz^2 = \lambda y \\
x^2y^2z = \lambda z \\
\implies \lambda x^2 = \lambda y^2 = \lambda z^2
$$
from this, either
$$
x^2 = y^2 = z^2
$$
leads to the maximum, or
$$
\lambda = 0 \implies xyz = 0
$$leads to the minimum.
A: So the equations you find by setting the derivatives equal to zero are
$$ 2x( y^2 z^2 + \lambda) = 0 \\
2y (x^2 z^2 + \lambda) = 0 \\
2z (y^2 x^2 + \lambda) = 0 \\
x^2 + y^2 + z^2 - r^2 = 0
 $$
Right, let's stop here and think. Suppose $x=0$. Then the first equation's obviously satisfied. The sphere equation can obviously be satisfied by various $y$ and $z$. The other two equations are
$$ 2y \lambda = 0 = 2z \lambda, $$
so can be satisfied by $\lambda=0$, whatever $y$ and $z$ are. Similarly starting from $y=0$ or $z=0$. If none of them are zero, you can cancel $x,y,z$ off the first 3 equations and carry on to find the other answers. Since $x^2y^2z^2$ is obviously at least zero, $x=0$ or $y=0$ or $z=0$ must give the minima (note these are not points, either, but circles $y^2+z^2=r^2$ and similar).
