Find extreme values for $f(x,y,z)=xyz$ given the constraints $g_{1}(x,y,z)=x+y+z-5$ and $g_{2}(x,y,z)=xy+yz+zx-8$ using Lagrange multipliers method. I want to calculate extreme values for  $f(x,y,z)=xyz$ given the constraints $g_{1}(x,y,z)=x+y+z-5$ and $g_{2}(x,y,z)=xy+yz+zx-8$ using Lagrange multipliers method. Im skeptical about my solution for this problem which goes as follow. First I got:
$$\nabla f= (yz)i+(xz)j+(xy)k,$$
$$\nabla g_{1}=i+j+k,$$
and
$$\nabla g_{2}=(y+z)i+(x+z)j+(y+x)k.$$
So from having the equality $\nabla f = \lambda_{1} \nabla g_{1} + \lambda_{2} \nabla g_{2}$ I got the following equation system:
\begin{align*}
yz&= \lambda_{1} + \lambda_{2}(y+z) \\
xz&=\lambda_{1}  + \lambda_{2}(x+z)\\
xy&=\lambda_{1} + \lambda_{2}(y+z)\\
\end{align*}
But Im really stucked finding the extrem values from the last equations system. So far,  I realize that if I sum up the three equations from the system and the way $g_{2}(x,y,z)$ is defined I obtained:
$$xy+yz+zx=8=3(\lambda_{1})+3(\lambda_{2})(x+y+z).$$
But from the way  $g_{1}(x,y,z)$ is defined I got that last equation is 
$$8=3(\lambda_{1})+3(\lambda_{2})(5).$$
So I found  by trial and error that $\lambda_{1}=1$ and $\lambda_{2}=\frac{1}{3}$, im not sure if there is more possible values for $\lambda_{1}$ and $\lambda_{2}$ satysfing the last equation. Anyways, from here I have been trying to find the values for $x,y$ and $z$ substituting the values I obtained for $\lambda_{1}$ and $\lambda_{2}$ in the equation from the original system of equations. For example, from the first equation I got 
$$yz=1+\frac{1}{3}(y+z)$$
But finding  $x,y$ and $z$ that way is hard. I´ve been thinking also that from some equation before I have that:
$$8=3\lambda_{1}+3\lambda_{2}(x+y+z)=3+(x+y+z)$$.
The problem of finding $x,y$ and $z$ this way is that I got a lot of points satisfying this last equation. Just to list a few:
$$(5,0,0),(0,5,0),(0,0,5),(1,1,3),...$$. Basically, all the points $(x,y,z) \in \mathbb{R}^{3}$ which satisfy $g_{1}(x,y,z)=0$ but I can tell there is something wrong from the fact a lot of these points dont satisfy the equations system obtained from $\nabla f = \lambda_{1} \nabla g_{1} + \lambda_{2} \nabla g_{2}$.
 A: Here $f(x,y,z) = xyz$,  $g(x, y, z) = x+y+z-5$, $h(x, y, z) = xy+yz+zx-8$
So $L(x, y, z, \lambda, \mu) = f(x, y, z) + \lambda g(x, y, z) + \mu h(x, y, z)$
$L_x = yz + \lambda +\mu(y+z) = 0$ --(1)
$L_y = xz + \lambda +\mu(x+z) = 0$ --(2)
$L_z = xy + \lambda +\mu(x+y) = 0$ --(3)
$L_{\lambda} = x+y+z = 5$ --(4)
$L_{\mu} = xy+yz+zx = 8$ --(5)
Multiply (1),(2),(3) by $x,y,z$ respectively, we get
$xyz +\lambda x +\mu(xy+xz) = 0$ --(6)
$xyz +\lambda y +\mu(xy+yz) = 0$ --(7)
$xyz +\lambda z +\mu(xz+yz) = 0$ --(8)
from (6) - (7) we get $(x-y)(\lambda+\mu z) = 0$ --(9)
from (7) - (8) we get $(y-z)(\lambda+\mu x) = 0$ --(10)
from (8) - (6) we get $(z-x)(\lambda+\mu y) = 0$ --(11)
Now if we take $x=y$ from (9), we see that it satisfies (10) and (11) also where $\lambda = -\mu x = -\mu y$, we are $\textbf{not}$ assuming here $z=x=y$
Then putting the vaue of x in (4) we get $2y+z = 5$, putting these values in (5) we get
$y^2 + 2y(5-2y) = 8$ or $3y^2 - 10y + 8 = 0$
Solving for y we get $y = 2,4/3$, so one solution is $2,2,1$ another solution is $4/3,4/3,7/3$
The solution $4/3,4/3,7/3$ gives the maximum value 4.148148, 
solution $2,2,1$ gives the minimum value 4. 
I am actually amazed to see such a narrow band!.
