System of 5x5 non-linear equations from Lagrange multiplier problem I have a Lagrange multiplier problem, and I've got down to the system of equations, but solving these has proved to be a nightmare for me.
I have five equations:
$$x+y+z=0$$
$$x^2+y^2+z^2=24$$
$$y=\lambda_1+2x\lambda_2$$
$$x=\lambda_1+2y\lambda_2$$
$$2=\lambda_1+2z\lambda_2$$
I keep getting down to $x=y$ but I know that a local minimum occurs when $x=\frac{1-3\sqrt5}{2},y=\frac{1+3\sqrt5}{2},z=-1$ and I can't seem to get down to that one.
I can see the answers, I'm just struggling with the working.
For reference, the original problem is to find the extrema for:
$$f(x,y,z)=xy+2z$$
with the constraints of $x+y+z=0$ and $x^2+y^2+z^2=24$
Please help...
 A: Subtracting your fourth equation from your third gives
$$ y-x=2\lambda_2(x-y)$$
 giving
$$ (2\lambda_2+1)(x-y)=0 $$
one solution of which is $\lambda_2=-\frac{1}{2}$.
Subtracting equation $5$ from equations $3$ and $4$ gives
\begin{eqnarray}
x-2&=&2\lambda_2(y-z)\\
y-2&=&2\lambda_2(x-z)
\end{eqnarray}
Using $\lambda_2=-\frac{1}{2}$ this gives
\begin{eqnarray}
x-2&=&z-y\\
y-2&=&z-x
\end{eqnarray}
that is, 
$$ x+y-z=2$$
Combining with your equation $1$ this gives 
$$ x+y=1$$
So $y=1-x$ and $z=-1$.
Substituting into your equation $2$ and simplifying gives
$$ x^2-x-11=0$$
which gives your missing solution.
A: One method is to eliminate the constraints. We have $x+y+z=0$ so $z=-x-y$. Also $x^2+y^2+z^2=1$ so
\begin{eqnarray*}
xy+yz+zx&=&-12.  \\
f =xy+2z &=& -12+z(2-x-y)=12+z(2+z)=(z+1)^2-13
\end{eqnarray*}
So the extreme point occurs at $z=-1$. Now lets calulate $x$ & $y$
\begin{eqnarray*}
x+y&=&1  \\
xy &=& -11  \\
(x-y)^2 &=& (x+y)^2-4xy=45
\end{eqnarray*}
So $x=\frac{1+3 \sqrt{5}}{2}$ & $y=\frac{1-3 \sqrt{5}}{2}$ (or vice versa).
