Maximizing $f$ in $\mathbb{R}^3$

Question

Find the domain and the maximum value that the function $$f(x,y,z)=\frac{x+2y+3z}{\sqrt{x^2+y^2+z^2}}$$ may attain in its domain.

I have found the domain of the function to be $\mathbb{R^3\backslash\mathbf{0}}$. To maximize I differentiated in terms of $x,y,z$ having $$f_x=\frac{-2 x y-3 x z+y^2+z^2}{\left(x^2+y^2+z^2\right)^{3/2}},\quad f_y=\frac{2 x^2-x y+z (2 z-3 y)}{\left(x^2+y^2+z^2\right)^{3/2}},\quad f_z=\frac{3 \left(x^2+y^2\right)-z (x+2 y)}{\left(x^2+y^2+z^2\right)^{3/2}}$$ But to solve the system $f_x=0,f_y=0$ and $f_z=0$ is rather hard. What are the plausible values of $x,y,z$?

Answer 1

Consider the vectors $\,\vec{u}=(x,y,z)\,$ and $\,\vec{v}=(1,2,3)$, then we can write $$\frac{x+2y+3z}{\sqrt{x^2+y^2+z^2}}=\frac{\vec{u}\boldsymbol{\cdot}\vec{v}}{\Vert\vec{u}\Vert}=\frac{\Vert\vec{u}\Vert\Vert\vec{v}\Vert\cos(\alpha)}{\Vert\vec{u}\Vert}=\Vert\vec{v}\Vert\cos(\alpha)=\sqrt{1^2+2^2+3^2}\cos(\alpha )=\sqrt{14}\cos(\alpha)$$

When is the last expression maximized? When $\alpha=k\pi,\,k\in\mathbb{Z}$, thus the maximum value of $f$ is $\sqrt{14}$.

Answer 2

By C-S $$(x^2+y^2+z^2)(1^2+2^2+3^2)\geq(x+2y+3z)^2,$$ which gives $$-\sqrt{14}\leq\frac{x+2y+3z}{\sqrt{x^2+y^2+z^2}}\leq\sqrt{14}.$$ The equality occurs for $(x,y,z)||(1,2,3),$ which gives that we got a maximal value and the minimal value.

Answer 3

Consider the line $x = t, y = 2t, z = 3t$

Along this line.

$f(t,2t,3t) = \frac {14 t}{\sqrt{14 t^2}} = \sqrt {14}$

let's find some orthogonal vectors.

$\mathbf u = (1,2,3)\\ \mathbf v = (2,-1,0)\\ \mathbf w = (0,3,-2)$

Any point in $\mathbb R^3$ is some linear combination

$c_1\mathbf u + c_2\mathbf v+ c_3\mathbf w$

$f(c_1\mathbf u + c_2\mathbf v+ c_3\mathbf w) = \frac {14c_1}{\sqrt{14c_1^2 + 5c_2^2 + 13c_3^2}}\\ |f|\le \sqrt 14 \text{ sgn}(c_1)$