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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?

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3 Answers 3

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Set $u=\sqrt2 \, x$, $v=\sqrt3 \, y$, $w=2 z$, so that you get a sphere $u^2+v^2+w^2=1$ instead of an ellipsoid. Are you able to use Cauchy–Schwarz now?

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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!

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Hint: write $$6x-3y-8z=3\sqrt{2}\cdot \sqrt{2}x-\sqrt{3}\cdot \sqrt{3}y-4\cdot2z$$

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