How Do You Use Lagrange Multipliers to Find the Closest Points to the Origin?

[Weatherkid11]

LaGrange Multipliers! Help!

Use the Lagrange multiplier method for 3 variables to find the points on the surface 3xy-z^2=1 that are closest to the origin.

I tried using the gradient= lamda(granient) and ended up getting (-3/2,0,-1). but i think i did it way wrong. Can someone please help? Thanks!

 

[mathwonk]

the gradient should be parallel to the radius vector, no?

 

[benorin]

Minimize $$f(x,y,z)=\sqrt{x^2+y^2+z^2}$$ subject to the constraint $$g(x,y,z)=3xy-z^2=1$$

and note that the min of f is also the min of f^2, so the vector eqn is

$$2<x,y,z> = \lambda <3y,3x,-2z>$$ and $$3xy-z^2=1$$

the z-component gives $$z=-\lambda z$$ and hence $$\lambda=-1$$ iff $$z\neq 0$$ and plugging $$\lambda=-1$$ into the vector eqn quickly gives both 2y=-3x and 2x=-3y so x=y=0, however this is an extraneous solution since (0,0,z) is not on the given suface (for real z anyhow). Therefore z=0 by contradiction and now it is easy.

Have fun,
-Ben