xyz^2=5 shortest distance

**flgator11** · May 26th 2008, 12:45 PM

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

Justify your solution.

Thanks alot!

**galactus** · May 26th 2008, 01:43 PM

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.

**Opalg** · May 27th 2008, 12:03 AM

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.

**galactus** · May 27th 2008, 04:31 AM

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

**flgator11** · May 27th 2008, 11:34 AM

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

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