# xyz^2=5 shortest distance

• May 26th 2008, 12:45 PM
flgator11
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!
• May 26th 2008, 01:43 PM
galactus
You could try a LaGrange multiplier.

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

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

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

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

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

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

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.
• May 27th 2008, 12:03 AM
Opalg
Quote:

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

In other words, we want to find the extrema of $\displaystyle f(x,y) = x^2+y^2+\frac5{xy}$. Put the two partial derivatives equal to zero and you get the equations $\displaystyle 2x=5/y^2$, $\displaystyle 2y=5/x^2$. Thus $\displaystyle 2x^4=5x$. Discounting the solution x=0, you see that $\displaystyle x=(5/2)^{1/3}$ (with y=x and z=2x). I think that's easier than using Lagrange multipliers for this question.
• May 27th 2008, 04:31 AM
galactus
Yes, it is easier, but Lagrange is a good thing to learn. (Clapping)
• May 27th 2008, 11:34 AM
flgator11
Well thanks to both of you for your help! It's much appreciated.