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):
- $\frac{x}{f}=\lambda yz^2$ or $\frac{x^2}{f}=2\lambda$
- $\frac{y}{f}=\lambda xz^2$ or $\frac{y^2}{f}=2\lambda$
- $\frac{z}{f}=2\lambda xyz$ 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.
Best Answer
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.