Question: . Use Lagrange multipliers to find the constrained critical points of f subject to the given constraints.
Here is the equation and the here is my solution. I am stuck now and I don't know how to proceed. I got a couple of restrictions but that's about it.
First of all: Nobody cares about the values of $\lambda$ and $\mu$. So get rid of them by first substituting $\lambda$ from (1) in (2) and (3). Subtract those two new equations to get $2\mu(-2y+3z)=0$. Convince yourself that $\mu=0$ yields to no solution. Hence $2y=3z$. Our first task is finished: no more $\lambda$, no more $\mu$. From here we easily arrive in $x=\pm\sqrt{13}/7$, $y=\mp18\sqrt{13}/91$ and $z=\mp12\sqrt{13}/91$.
That gives a minimum and a maximum as expected, since the first constraint is a plane and the second the unit sphere and we're searching for those points with extremal first coordinate.
