# Homework Help: Why does the Hessian determinant Δ_p = -1 imply that P(0, 0) is a saddle point?

1. Apr 16, 2012

### s3a

1. The problem statement, all variables and given/known data
For the function f(x, y) = xye^[-(x^2 + y^2)] find all the critical points and classify them each as a relative maximum, a relative minimum, or a saddle point.

2. Relevant equations
Partial differentiation and Hessian determinants.

3. The attempt at a solution
I get how to compute the derivatives. I also get how to compute the Hessian determinants. Basically, I get all the algebraic details but what I would like to ask about (at least for now) is why does the Hessian determinant Δ_p = -1 imply that P(0, 0) is a saddle point?

Any input would be greatly appreciated!

#### Attached Files:

• ###### Solution.jpg
File size:
41.5 KB
Views:
112
2. Apr 16, 2012

### Ray Vickson

The Hessian at (x0,y0) gives the quadratic terms in the Taylor expansion at that point. In other words, if we write
$$f(x,y) \approx f(x_0,y_0) + a_0 (x-x_0) + b_0 (y-y_0) + \frac{1}{2} r_0 (x-x_0)^2 + s_0 (x-x_0)(y-y_0) + \frac{1}{2} t_0 (y-y_0)^2 + \text{ higher order terms },$$ the constants $r_0, s_0, t_0$ are the elements of the Hessian at (x0,y0); that is, the Hessian is
$$H(x_0,y_0) = \left[ \matrix{r_0&s_0\\s_0&t_0} \right].$$ So, if the Hessian is $$H = \left[ \matrix{0&1\\1&0}\right],$$ that means that for small x and y we have $f(x,y) \approx xy,$ and if you plot the surface z = xy around (0,0), you will see that (0,0) is, indeed, a saddle point.