The global maximum and the global minimum of the function $f(x,y,z)=xyz$ with the constraint $x^2+2y^2+3z^2=6$ can be found using Lagrange multipliers.
$\nabla f = \lambda \nabla g$
$g(x,y,z)=x^2+2y^2+3z^2-6=0$
This results into:
$$yz = \lambda 2x$$
$$xz = \lambda 4y$$
$$xy=\lambda6z $$
When looking at the equation $x^2+2y^2+3z^2-6=0$, it can be seen that there are six solutions $(x,y,z)$ with two coordinates equal to zero: $(0,0,\pm \sqrt 2 ), (0, \pm \sqrt 3, 0 ) , (\pm \sqrt 6, 0, 0 ) $
But I got stuck here. Could someone help me with this?
Best Answer
You should try to solve the equations. The six solutions wouldn't help. Following @Mann's idea, you can multiply the left hand sides of the three equations, and multiply the right hand sides of the three equations, then equate them:
$$xy\cdot yz\cdot xz =48 \lambda^3 xyz$$
then see if you can solve the $x,y,z$'s with the obtained information.
To make the solution complete, you should first observe that any one of $x,y,z=0$ is impossible, since one of them will imply the others and that contradict with your constraint.
So you get $xyz=48\lambda^3$. Notice your three original equations are in a pattern that is very consistent with this. For example, you can divide by $x$ to get $yz=\frac{48\lambda^3}{x}$ and plug this into the first equation. This gives you $x^2=24\lambda^2$. Doing the same thing for the other two equations, you can get similar equations for $y^2,z^2$. Now plug these into the constraint you will get an equation with respect to $\lambda^2$. Can you solve it from here?