Finding the minimum distance from the origin to the surface $xyz^2=2$

Question

This was an old exam question I was looking at for a friend, although it's been a while since I've done this stuff:

Q. Find the shortest distance from the origin to the surface $xyz^2=2$.

I remembered Lagrange Multipliers being used in that course, so I did the following.

Define $f(x,y,z)=\sqrt{x^2+y^2+z^2}$, which is what we want to minimize, with the constraint $g(x,y,z)=xyz^2=2$.

We get three equations (I'm using $f$ as shorthand to avoid the messy root):

1. $\frac{x}{f}=\lambda yz^2$ or $\frac{x^2}{f}=2\lambda$
2. $\frac{y}{f}=\lambda xz^2$ or $\frac{y^2}{f}=2\lambda$
3. $\frac{z}{f}=2\lambda xyz$ or $\frac{z^2}{f}=4\lambda$

We get $y=\pm x$, $z=\pm \sqrt{2} \cdot x$ and with the constraint only $y=x$ will work.

Finally we get $x=\pm 1$ and this gives us $4$ points $\pm(1,1,\sqrt2)$ and $\pm(1,1,-\sqrt2)$.

I haven't really used $\lambda$ though... so it's like I didn't really do it properly.

Answer 1

No one can be zero. So, in that surface, $z^2=2/xy$. Now, by AM-GM

$$x^2+y^2+z^2=x^2+y^2+\frac{2}{xy}\geq2xy +\frac{2}{xy}\geq 2\sqrt{4}=4.$$ Study the conditions for the equality to happen and show that they actually happen for the points you already suspect.

Answer 2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ Use $\ds{\tt\mbox{Spherical Coordinates}}$ and a $\ds{\tt\mbox{Lagrange Multiplier}}$ $\ds{4\mu}$:

\begin{align} {\cal F}&\equiv r - 4\mu\braces{\bracks{r\sin\pars{\theta}\cos\pars{\phi}} \bracks{r\sin\pars{\theta}\sin\pars{\phi}} \bracks{r\cos\pars{\theta}}^{2} -2} \\[3mm]&=r-4\mu\bracks{r^{4}\sin^{2}\pars{\theta}\cos^{2}\pars{\theta}\sin\pars{\phi} \cos\pars{\phi} - 2} \end{align}

$$ {\cal F}=r - \half\,\mu r^{4}\sin^{2}\pars{2\theta}\sin\pars{2\phi} + 8\mu $$

$$ \begin{array}{rclcrcl} \partiald{{\cal F}}{r} & = & 0 & \imp & 1 - 2\mu r^{3}\sin^{2}\pars{2\theta}\sin\pars{2\phi} & = & 0 \\[2mm] \partiald{{\cal F}}{\theta} & = & 0 & \imp & -\mu r^{4}\sin\pars{4\theta}\sin\pars{2\phi} & = & 0 \\[2mm] \partiald{{\cal F}}{\phi} & = & 0 & \imp & -\mu r^{4}\sin^{2}\pars{2\theta}\cos\pars{2\phi} & = & 0 \end{array} $$

$\ds{\theta \not\in\braces{0,\pi}}$ and $\ds{\phi \not\in\braces{0,\pi,2\pi}}$. That leads to $\ds{\theta =\phi = {\pi \over 4}}$: $$ 2=xyz^{2}={1 \over 8}\,r^{4}\sin^{2}\pars{\pi \over 2}\sin\pars{\pi \over 2} ={r^{4} \over 8}\ \imp\ \begin{array}{|c|}\hline\\ \quad\color{#66f}{\Large r = 2}\quad \\ \\ \hline \end{array} $$

Answer 3

Here is another AM-GM based approach:

$\frac{x^2 + y^2 + \frac{z^2}{2} + \frac{z^2}{2}}{4} \ge \sqrt[4]{ \frac{x^2y^2z^4}{4}} = 1$

So the square of the distance is at least $4$, with equality iff $x^2 = y^2 = \frac{z^2}{2} = 1$. Taking signs into account gives you the correct solution set.