Second Derivative Test

1. Aug 9, 2013

Zyuke

Under what conditions on the constants a and b does the second derivative test guarantee that the function

g(x,y,z)=ax^2+2axz+by^2-2byz+z^2

has a local maximum at (0,0,0)? a local minimum at (0,0,0)?

well, i used the Hessian matrix to compute the eigenvalues to set them above zero. but the computation is so complicated that i used Mathematica to solve them and it turned out to be some messy stuff. i was convinced that this is not the way to do it. but then how?

2. Aug 9, 2013

Jufro

[STRIKE]Just looking at the problem makes me want to try Lagrange multipliers.[/STRIKE]

Last edited: Aug 9, 2013
3. Aug 9, 2013

micromass

Staff Emeritus
Won't work here.

They ask about the second derivative test, so you'll need to calculate the Hessian. It's what they ask for.

4. Aug 9, 2013

Ray Vickson

Why would you try to set the eigenvalues to zero? The Hessian is positive-definite if all the eigenvalues are > 0 (so the origin is a global minimum---not just local!)

The Hessian cannot be negative-definite, so the origin cannot be a local maximum. (Note: a matrix A is negative-definite if and only if its negative B = -A is positive-definite, and since B(3,3) = -2 < 0, B cannot be positive-definite---no fancy tests needed here!)

To test for positive-definiteness of the Hessian H, just perform Cholesky decomposition; see, eg., http://en.wikipedia.org/wiki/Cholesky_decomposition . The Cholesky algorithm is easy to carry out in this case.

5. Aug 10, 2013

Zyuke

thanks a lot.