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.
 A: Hint: CS inequality $\implies$
$$(2x^2+3y^2+4z^2)(8+3+1) \ge (4x-3y-2z)^2$$
A: Try this: Compute the gradient of $$\vec\nabla S = \begin{pmatrix}\frac{\partial S}{\partial x}\\\frac{\partial S}{\partial y}\\\frac{\partial S}{\partial z}\end{pmatrix} =\begin{pmatrix}4\\-3\\-2\end{pmatrix} $$
Now find all the points where this gradient is prependicular to your surface defined by $T(x,y,z)= 2x^2 + 3y^2 + 4z^2 = 1$
To do so compute the gradient of your surface function $T$
$$\vec\nabla T(x,y,z) = \begin{pmatrix}\frac{\partial T}{\partial x}\\\frac{\partial T}{\partial y}\\\frac{\partial T}{\partial z}\end{pmatrix} =\begin{pmatrix}4x\\6y\\8z\end{pmatrix} $$
Then you get $x=\lambda$ $y=-2\lambda$ and $z=-2\lambda$
Find now the two values $\lambda$ such that $T(\lambda)=1$
$$T(\lambda)=2\lambda^2+12\lambda^2+16\lambda^2=30\lambda^2=1$$
This means $\lambda_{1,2}=\pm\sqrt{1/30}$
This leads to $S(\lambda_1)=7\sqrt{\frac{2}{15}}$ and $S(\lambda_2)=-7\sqrt{\frac{2}{15}}$. So have the maximum and minimum of $S(x,y,z)$ while the condition $T(x,y,z)=1$
[Edit]
I think I messed up.
