$2x^2+ 3y^2+4z^2 =1$ find the maximum of $4x+3y+2z$ If $2x^2+ 3y^2+4z^2 =1$ find the maximum of $4x+3y+2z$.
This is a question from a regional math olympiad and thus there must exist solutions without application of calculus.
I have no idea how to begin.
 A: Apply Cauchy-Schwarz Inequality: $\left(4x+3y+2z\right)^2 = \left(2\sqrt{2}\cdot \sqrt{2}x+\sqrt{3}\cdot \sqrt{3}y+1\cdot 2z\right)^2 \leq \left((2\sqrt{2})^2+(\sqrt{3})^2+1^2\right)\cdot\left(2x^2+3y^2+4z^2\right) = 12 \Rightarrow 4x+3y+2z \leq 2\sqrt{3} = \text{Maximum Value}$.
A: For me,
$4x + 3y + 2z = \langle\begin{pmatrix} \sqrt{2} x \\ \sqrt{3}y \\ 2z\end{pmatrix}, \begin{pmatrix} 2\sqrt{2} \\\sqrt{3} \\\ 1\end{pmatrix}\rangle \leq 1\times 2\sqrt{3}=2\sqrt{3}$
A: Just another way: sum the AM-GMs...
$$2x^2+2k^2 \ge k\cdot 4 x,\quad 3y^2+\frac34k^2 \ge k \cdot 3 y, \quad 4z^2+ \frac14k^2 \ge k\cdot2z$$
to get $\dfrac1k+3k \ge (4x+3y+2z)$, and the maximum is achieved when $\frac1k=3k\implies k = \frac1{\sqrt3}$, for a maximum of $2\sqrt3$. 
A: First, change the ellipsoid to a sphere by writing $v = x \sqrt{2}, w=y\sqrt{3}, u = 2z$
THe function to be maximized will still be linear, now in $u, v, w$.
$$ a_u u + a_v v + a_w w $$
Now proceed in the direction perpendicular to that plane, and lets look at a lihne thru the origin in that direction, which could be parameterized as
$$
u = s/a_u \\ v = s/a_v \\ w = s/a_w
$$
The maximizing point will be at the value of $s$ where the line meets the sphere, and this is now easy algebra.
A: WLOG, we can set $\sqrt2x=\cos A,\sqrt3y=\sin A\cos B,2z=\sin A\sin B$
$\implies4x+3y+2z=2\sqrt2\cos A+\sqrt3\sin A\cos B+\sin A\sin B$
$=2\sqrt2\cos A+\sin A(\sqrt3\cos B+\sin B)$
$=2\sqrt2\cos A+2\sin A\cos\left(B-\dfrac\pi6\right)$
If $\sin A\ge0,$
$4x+3y+2z\le2\sqrt2\cos A+2\sin A$ as $\cos\left(B-\dfrac\pi6\right)\le1$
The equality occurs if $\cos\left(B-\dfrac\pi6\right)=1$
$\iff B-\dfrac\pi6=2m\pi\iff B=2m\pi+\dfrac\pi6$ where $m$ is any integer
Again, $2\sqrt2\cos A+2\sin A=2(\sqrt2\cos A+\sin A)=2\sqrt3\sin\left(A+\arcsin\dfrac1{\sqrt3}\right)\le2\sqrt3$
The equality occurs if $\sin\left(A+\arcsin\dfrac1{\sqrt3}\right)=1$
$\iff A+\arcsin\dfrac1{\sqrt3}=2n\pi+\dfrac\pi2$
$\iff A=2n\pi+\dfrac\pi2-\arcsin\dfrac1{\sqrt3}=2n\pi+\arccos\dfrac1{\sqrt3}$
 where $n$ is any integer
Similarly, if $\sin A<0,$
$4x+3y+2z\le2\sqrt2\cos A+2\sin A$ as $\cos\left(B-\dfrac\pi6\right)\ge-1$
A: It is obvious that the first equation stands closed surface, so when 
$4x+3y+2z=k$ gets the minimum, the line just touches the closed surface. Let me put the tangent point $A(x_0，y_0)$, and then I get the partial derivative of  $2x_2+3y_2+4z_2=1$ about $X$ and $Y$ at the point $A$. I know I can find a line that coincides with the given line. After this I can get $x_0=2y_0$, and $x_0,y_0$ also satisfy the equation $2x_2+3y_2+4z_2=1$ and $4x+3y+2z=k$.
I can then get the minimum of $k$ easily.
