1. ## xyz^2=5 shortest distance

Given the surface xyz^2=5. Prove that on this surface there exists a point closest to the origin, and find that point.

Thanks alot!

2. You could try a LaGrange multiplier.

We want to find the extrema of $f(x,y,z)=x^{2}+y^{2}+z^{2}$ subject to the constraint $xyz^{2}=5$

Therefore, we have $g(x,y,z)=xyz^{2}$

${\nabla}f(x,y,z)={\lambda}{\nabla}g(x,y,z)$

$xi+yj+zk={\lambda}(yz^{2}i+xz^{2}j+2zxyk)$

$x={\lambda}yz^{2}; \;\ y={\lambda}xz^{2}; \;\ z={\lambda}2zxy$

${\lambda}=\frac{x}{yz^{2}}...[1]; \;\ {\lambda}=\frac{y}{xz^{2}}...[2]; \;\ {\lambda}=\frac{1}{2xy}...[3]$

Now, you can set [1] and [2] equal and solve for y. Set [1] and [3] equal and solve for z. Then sub them into the constraint to find your x value. Then you can sub that back into the other equations to find y and z.
Plug those into f(x,y,z) to find the point that is closest the origin.
There may be several to check.

3. Originally Posted by galactus
We want to find the extrema of $f(x,y,z)=x^{2}+y^{2}+z^{2}$ subject to the constraint $xyz^{2}=5$
In other words, we want to find the extrema of $f(x,y) = x^2+y^2+\frac5{xy}$. Put the two partial derivatives equal to zero and you get the equations $2x=5/y^2$, $2y=5/x^2$. Thus $2x^4=5x$. Discounting the solution x=0, you see that $x=(5/2)^{1/3}$ (with y=x and z=2x). I think that's easier than using Lagrange multipliers for this question.

4. Yes, it is easier, but Lagrange is a good thing to learn.

5. Well thanks to both of you for your help! It's much appreciated.