<s>[INST] Find the maximum value of $4x − 3y − 2z$ subject to $2x^2 + 3y^2 + 4z^2 = 1.$ Find the maximum value of $4x − 3y − 2z$ subject to $2x^2 + 3y^2 + 4z^2 = 1.$
My Attempt
let $S=4x − 3y − 2z$ and $ t=2x^2 + 3y^2 + 4z^2$. Then $t-s =2x^2 + 3y^2 + 4z^2 -(4x − 3y − 2z)= 2(x-1)^2 + 3(y+1/2)^2+4(z+1/4)^2 -3>=-3$
so $s<=t+3<=4$ as $t=1$ So minimum value of S is 4.
It my reasoning and answer correct? I am novice in this type of problems and not confident that I have arrived a right solution or not. Please help.
 [/INST] Hint: CS inequality $\implies$
$$(2x^2+3y^2+4z^2)(8+3+1) \ge (4x-3y-2z)^2$$
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