<s>[INST] 3 variable multiplication with 1 constraint lagrange multiplier Using Lagrange multipliers, I need to calculate all points $(x,y,z)$ such that $$x^4y^6z^2$$ has a maximum or a minimum subject to the constraint that $$x^2 + y^2 + z^2 = 1$$
So, $f(x,y,z) = x^4y^6z^2 $
and $g(x,y,z) = x^2 + y^2 + z^2 - 1$
then i've done the partial derivatives
$$\frac{\partial f}{\partial x}(x,y,z)=\lambda\frac{\partial g}{\partial x}$$
which gives
$$4x^3y^6z^2 = 2xλ$$
$$6x^4y^5z^2 = 2yλ$$
$$2x^4y^6z = 2zλ$$
which i subsequently go on to find that 
$3x^2 = 2y^2 = 6z^2 $
This is where i've hit a dead end. Where do i go from here? or am i doing it all wrong? 
Thanks.
 [/INST] The minimal value is $0$ for $x=0$.
The maximal value we can find by AM-GM:
$$x^4y^6z^2=108\left(\frac{x^2}{2}\right)^2\left(\frac{y^2}{3}\right)^3z^2\leq108\left(\frac{2\cdot\frac{x^2}{2}+3\cdot\frac{y^2}{3}+z^2}{6}\right)^6=\frac{1}{432}.$$
The equality occurs for $\frac{x^2}{2}=\frac{y^2}{3}=z^2$ and $x^2+y^2+z^2=1$,
which says that $\frac{1}{432}$ is a maximal value.
Done!
If you wish to use the Lagrange multipliers method you need to add the following words.
Let $F(x,y,z,\lambda)=x^4y^6z^2+\lambda(x^2+y^2+z^2-1)$ and $A=\left\{(x,y,z,\lambda)|x^2+y^2+z^2=1\right\}$.
Sinse $F$ is a continuous function and $A$ is a compact, we see that $F$ gets on $A$ the maximal value and gets on $A$ the minimal value, which happens for solutions of your system.
The rest is to solve the system $\frac{x^2}{2}=\frac{y^2}{3}=z^2$ and $x^2+y^2+z^2=1$ for $xyz\neq0$ and to solve your system for $xyz=0$.
Now, you can choose, that you want. 
I think that to solve our problem by AM-GM is much better. 
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