System of 5x5 non-linear equations from Lagrange multiplier problem

Question

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...

Answer 1

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.

Answer 2

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).