TeX loves LATEXDoes F=x2y2-2x-2y have a minimum at the point x=y=1?

[tatianaiistb]

Homework Statement

Decide whether F=x²y²-2x-2y has a minimum at the point x=y=1 (after showing that the first derivatives are zero at that point).

Homework Equations

F_xxF_yy-F_xy²

The Attempt at a Solution

So I found that:
F_x=2xy²-2, which at point (1,1) = 0 OK
F_y=2x²y-2, which at point (1,1) = 0 OK
F_xx=2y²
F_yy=2x²
F_xy=4xy

So at point (1,1):
F_xxF_yy-F_xy²=4-16=-12, which is less than 0.

Does this mean that because it is less than zero, rather than greater, it DOES NOT have a minimum but rather a saddle point? Thanks!

 

[I like Serena]

Hi tatianaiistb!

Yes, the second derivative test says that a negative determinant means it's a saddle point.

Btw, you can easily verify by checking a few points in the neighbourhood, like (1,0) and (-1.1).

 

[Ray Vickson]

Homework Statement

Decide whether F=x²y²-2x-2y has a minimum at the point x=y=1 (after showing that the first derivatives are zero at that point).

Homework Equations

F_xxF_yy-F_xy²

The Attempt at a Solution

So I found that:
F_x=2xy²-2, which at point (1,1) = 0 OK
F_y=2x²y-2, which at point (1,1) = 0 OK
F_xx=2y²
F_yy=2x²
F_xy=4xy

So at point (1,1):
F_xxF_yy-F_xy²=4-16=-12, which is less than 0.

Does this mean that because it is less than zero, rather than greater, it DOES NOT have a minimum but rather a saddle point? Thanks!

You get a saddle point. To understand WHY (rather than just citing prescriptions), look at what you get when you use the Hessian matrix to construct a quadratic approximation near (1,1): f ~ 2(x-1)^2 + 2(y-1)^2 -8(x-1)(y-1) = 2[(x-1)-(y-1)]^2-4(x-1)(y-1) = 2(x-y)^2 -4(x-1)(y-1) = 2(u-v)^2-4uv, where u=x-1 and v=y-1. When v=0 we have f ~ 2u^2 > 0 for nonzero u; when u=0 we have f~2v^2 > 0 for nonzero v. But when u=v we have f ~ -4u^2 < 0 for nonzero u (and v), so (u,v) = (0,0) is a min along some direction lines and a max along other direction lines. That is what a saddle point means.

RGV