TITLE: Solving without using of Lagrange multipliers
QUESTION [5 upvotes]: Let $M$ be the maximum value of $6x - 3y - 8z$ subject to $2x^2 + 3y^2 +4z^2 = 1$

I have applied CS and AM-GM but am not able to kill the problem. For the test I am preparing for, permits not the use of Lagrange multipliers. So is there any other way?

REPLY [7 votes]: By C-S
$$6x-3y-8z\leq|6x-3y-8z|\leq\sqrt{(18+3+16)(2x^2+3y^2+4z^2)}=\sqrt37.$$
The equality occurs for $(\sqrt{18},\sqrt3,4)||(\sqrt2x,-\sqrt3y,-2z)$,
which says that $\sqrt37$ is indeed a maximal value.
Done!