While trying to work on a Lagrange multiplier problem, I encountered a system of linear equations that I'm not really able to solve. I don't know where to start. I've already found that $x=y$ or $\mu=-0.5$. I need to find the stationary points given the following set of equations:
$$x+y-z=0$$ $$x^2+y^2+z^2-6=0$$ $$y-\lambda-2\mu x=0$$ $$x-\lambda-2\mu y=0$$ $$2+\lambda-2\mu z=0$$
The original problem is the following:
Find the points where $f(x,y,z)=xy+2z$ is stationary subject to $x+y-z=0$ and $x^2+y^2+z^2=6$.
A science student assistant was already unable to help me. I'm very sorry for maybe not following the standards set on this website, but I'm new to stack exchange.
Let's order the equations as 1, 2, 3, 4, 5. You probably get $x=y$ or $\mu=-0.5$ by plugging 3 into 4 and some manipulation. And that's correct.
Now if $u=-0.5$, plug it into 3, 4, 5, you will get
$$x+y-\lambda =0\\ 2+\lambda +z =0$$
Since 1 gives you $x+y=z$, you can get $\lambda =x+y=z $ and $\lambda=-z-2$. Combining these gives you $z=-1$. Then use $x+y=z=-1$ and 2, you can get $(1, -2, -1)$ and $(-2, 1 , -1)$.
If $x=y$, from 1 you also have $x+y=z$, so $z=2x$. Plugging $y=x$ and $z=2x$ into 2, you can get $x=\pm 1$. I think you can continue from there.