1. ## Lagrange multipliers

What points on the the curve $x^2+2xy+2y^2= 100$ are closest to the origin? I think that $F(x,y,z,\lambda)=x^2+y^2+z^2+\lambda(x^2+2xy+2y^2-100)$
Then $F_{x}=2x+2\lambda x+2\lambda y$
$F_{y}=2y+2\lambda x+4\lambda y$
$F_{z}=0$
$F_{\lambda}=x^2+2xy+2y^2-100$
If I set those equal to zero I get
$2x=-2\lambda x-2\lambda y$
$2y=-2\lambda x-4\lambda y$
$x^2+2xy+2y^2=100$
I'm having trouble solving that system. I don't know if I've got the partials wrong, or am just doing the algebra wrong. Any helpful thoughts?
Thanks

2. ## Re: Lagrange multipliers

partials are correct. Solve for x in the first equation.
Note the 3rd equation is

$$(x+y)^2=100 \Rightarrow (x+y)=\pm 10$$

So plug what you have for x into these two eqs and solve for y.

Then solve for lambda in the final eq.

There are two minima and 2 maxima.