Method to find the extremal values of $xyz$ subject to $x^2+2y^2+3z^2=a$ This question has been asked before but I want to lay out my method and get feedback on reasoning and process this took me a long to put together as I am new to the formatting: 
Let the function $f$ be defined as
$f$($x$,$y$,$z$) $=$ $x$$y$$z$
find the maximum and minimum values subject to the constraint:
$g$($x$,$y$,$z$) $=$ $x^2$+2$y^2$+3$z^2$ 
Equation 1 $=$ $\nabla$$f_x$ $=$ $\lambda$ $\nabla$$g_x$ $=$ [ $yz$ $=$ $\lambda$2$x$]
Equation 2 $=$ $\nabla$$f_y$ $=$ $\lambda$ $\nabla$$g_y$ $=$ [ $xz$ $=$ $\lambda$4$y$]
Equation 3 $=$ $\nabla$$f_z$ $=$ $\lambda$ $\nabla$$g_z$ $=$ [ $yx$ $=$ $\lambda$6$z$]
It is my understanding that to solve these equations really just amounts to solving the relationship between values $x,y,z$ (and then plug into constraint) or the values themselves if there is an easy way to do so.
My basic instinct was to solve Equation 1 for $x$, obtaining:
$x$ $=$ $\frac{yz}{2\lambda}$
Since Equation 3 consists of $x,y,z$ and I have x in terms of $y$ and $z$, then plugging $x$ into Equation 3 would get me the relationship of $y$ to $z$, giving me:
$\frac{yz}{2\lambda}$$\frac{y}{1}$ $=$ $\frac{\lambda6z}{1}$
Simplifying I then get:
12$z$$\lambda^2$ $=$ $y^2$$z$
Then Subtracting the right side over to the left and factoring out $z$ gives:
$z$(12$\lambda^2$-$y^2$) $=$ 0
Giving me:
$z$ $=$ 0 and 12$\lambda^2$ $-$ $y^2$ $=$ 0
$\lambda$ $\neq$ 0 because that would imply that $x=y=z=0$ which does not hold true in the constraint.
Now following from the above observation $z$ $\neq$ 0 because that would imply that $x=y=z=0$ which does not hold true in the constraint. This leaves 12$\lambda^2$ $-$ $y^2$ $=$ 0 for me to obtain values from. 
Solving for $y$ I get:
$y$ $=$ $\pm$ 2$\lambda$$\sqrt{3}$
I then plugged $y$ into the $x$ value I solved for in my first step: $x$ $=$ $\frac{yz}{2\lambda}$ and obtained:
$x$ $=$ $\pm$ $\sqrt{3}z$
From here I am lost. I can't seem to do anything to get the rest of the values and get numerical values. I would like my process addressed first and then suggestions, and by that I mean answering the question is there a way to proceed down the same route that I am going or was my process flawed from the start? The book gives the answer exactly as follows:  maximum $\frac{2}{\sqrt{3}}$, minimum $\frac{-2}{\sqrt{3}}$. What does that even mean when checking my answers I am used to $f$($x,y,z$) $=$ $c$. I'm guessing that in this instance they are just giving me c. 
Thank you 
 A: I have not been able to find the original question and I give you here my approach hoping it will help you.
We want to maiximize $xyz$ subject to the constraint $x^2+2 y^2+3 z^2=a$ with ($x\geq 0$) , ($y\geq 0$) , ($z \geq 0$). 
So, let us consider $$F=x y z +\lambda  \left(x^2+2 y^2+3 z^2-a\right)$$ Computing derivatives $$F'_x=y z+2 \lambda  x=0\tag 1$$ $$F'_y=x z+4 \lambda  y=0\tag 2$$ $$F'_z=x y+6 \lambda  z=0\tag 3$$ $$F'_\lambda=x^2+2 y^2+3 z^2-a=0\tag 4$$ Now, I should consider equations $(1,2,3)$ and solve them for $x,y,z$ in terms of $\lambda$. 
Using $(1)$ gives $x=-\frac{y z}{2 \lambda }$; plugging in $(2)$ leads to $y \left(4 \lambda -\frac{z^2}{2 \lambda }\right)=0$;  plugging in $(3)$ leads to
$z\left(6 \lambda -\frac{y^2}{2 \lambda }\right)=0$.
Considering all possibilities, the solutions of equations $(1,2,3)$ are then (hoping no mistakes)
$$x=0\qquad y= 0\qquad z= 0$$
$$x= -2 \sqrt{6} \lambda \qquad y= -2 \sqrt{3} \lambda \qquad z=-2 \sqrt{2} \lambda$$
$$x= -2 \sqrt{6} \lambda \qquad y= +2 \sqrt{3} \lambda \qquad z=+ 2 \sqrt{2} \lambda$$
$$x= +2 \sqrt{6} \lambda \qquad y= -2 \sqrt{3} \lambda \qquad z= +2 \sqrt{2} \lambda$$
$$x= +2 \sqrt{6} \lambda \qquad y=+2 \sqrt{3} \lambda \qquad z= -2 \sqrt{2} \lambda$$  Using $(4)$, we just get $72 \lambda^2=a$, then $\lambda$ and the remaining follows.
