Mesmerizing Maxima

By Cauchy-Schwarz inequality, we get

$\left\{ {(\sqrt{2}x)}^{2}+{(\sqrt{3}y)}^{2}+{(2z)}^{2} \right\} \left\{ {(3\sqrt{2})}^{2}+{(-\sqrt{3})}^{2}+{(-4)}^{2} \right\} \geq {(6x-3y-8z)}^{2}$

Therefore, $M=\sqrt{37}$ and the answer is $6$ .

The maximum value of $6x-3y-8z$ must occur when $x\geq0, y\leq0, z \leq0$ . Let $(X,Y,Z) = (x,-y,-z)$ . Then we want to maximize $6X + 3Y + 8Z$ subject to the constraints $2X^2 + 3Y^2 + 4Z^2 = 1$ for non-negative numbers $X,Y,Z$ .

Let $N = 6X + 3Y + 8Z$ , then $3Y = N - 6X - 8Z \quad \Rightarrow \quad 9Y^2 = (6X + 8Z - N)^2 .$

Substitute this value into the constraint $2X^2 + 3Y^2 + 4Z^2 = 1 \quad\Leftrightarrow\quad 6X^2 + 9Y^2 + 12Z^2 = 3$ gives $6X^2 + (6X + 8Z - N)^2 + 12Z^2 = 3.$

When $N$ is maximized, this means that the plane $N = 6X + 3Y + 8Z$ touches the ellipsoid $2X^2 + 3Y^2 + 4Z^2 = 1$ at exactly 1 point. In other words, the (quadratic) discriminant of the equation above (in $X$ , and in $Z$ ) are both 0 ( $B^2-4AC=0$ ):

$\begin{cases} 12^2 (8Z - N)^2 - 4 \cdot 72 \cdot(-N^2 - 3) = 0 \\ 16^2 (6X - N)^2 - 4 \cdot 76 \cdot (-N^2 - 3) = 0 \end{cases}$

We have formed another two quadratic equations, both have a discriminant of 0 too!

This gives $\max(N) = \sqrt{37}$ . Hence $M = \max(N) = \sqrt{37} \Rightarrow \lfloor S \rfloor = \boxed6$ .

But when does this maximum value occur? It occurs when $N = \sqrt{37} \quad \Rightarrow \quad (X,Z) = \left( \dfrac{3}{\sqrt{37}}, \dfrac2{\sqrt{37}} \right) \quad \Rightarrow \quad Y = \dfrac1{\sqrt{37}} \quad \Rightarrow \quad (x,y,z) = \left( \dfrac3{\sqrt{37}}, -\dfrac1{\sqrt{37}}, -\dfrac2{\sqrt{37}} \right) .$

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Alternative solution via calculus : We will solve this via Lagrange multipliers (LM).

Let $f(x,y,z) = 2x^2 + 3y^2 + 4z^2$ and $g(x,y,z) = 6x-3y-8z$ . By LM, we have $\Delta f = \lambda \Delta g \quad \Rightarrow \quad \langle 4x, 6y,8z\rangle = \lambda\langle 6,-3,8 \rangle .$

This means that $\lambda = \dfrac{4x}6 = \dfrac{6y}{-3} = \dfrac{8z}{-8} \quad \Rightarrow \lambda = \pm \sqrt{37}$

Thus, $M = \max(\lambda) = \sqrt{37} \Rightarrow \lfloor S \rfloor = \boxed6$ . Equality holds when $\dfrac{2x}3 = -2y = -z \quad\Leftrightarrow\quad (x,y,z) = \left( \dfrac3{\sqrt{37}}, -\dfrac1{\sqrt{37}}, -\dfrac2{\sqrt{37}} \right)$ .

But is this expression maximized? Just use the Hessian matrix on $f(x,y,z)$ to get a determinant $\det { \begin{bmatrix}{\tfrac{\partial^2 f}{\partial x^2} } && {\tfrac{\partial^2 f}{\partial y \partial x} } && {\tfrac{\partial^2 f}{\partial z \partial x} } \\ {\tfrac{\partial^2 f}{\partial x\partial y} } && {\tfrac{\partial^2 f}{\partial y^2} } && {\tfrac{\partial^2 f}{\partial z \partial y} } \\ {\tfrac{\partial^2 f}{\partial x \partial z} } && {\tfrac{\partial^2 f}{\partial y \partial z} } && {\tfrac{\partial^2 f}{\partial z^2} }\end{bmatrix} } = \det { \begin{bmatrix}{4} && {0} && {0} \\ {0} && {6} && {0} \\ {0} && {0} && {8}\end{bmatrix} } = 4\cdot 6\cdot8 > 0$ and we're done!