# Function in 3 variables, determinant of the Hessian=0

1. Nov 16, 2013

### Felafel

1. The problem statement, all variables and given/known data

find the minima and maxima of the following function:
$f:\mathbb{R}^3 \to \mathbb{R} : f(x,y,z)=x(z^2+y^2)-yx$

3. The attempt at a solution

after computing the partials, i see ∇f=0 for every point in the x-axis: (a, 0, 0)
The Hessian is:
( 0 0 0 )
( 0 2a -1 )
( 0 -1 2a )
for every value of a, the determinant is 0.
the eigenvalues are: $\lambda_1=0$
$\lambda_{2,3}= 2a±\sqrt{1}$=$2a\pm1$
$\lambda_{2,3}$ are both positive if a>1/2 and both negative if a<-1/2.
Thus, for -1/2<a<1/2 i can say the points are inflection points, because the eigenvalues have oppisite sign.
But how about $a \geq 1/2$ and $a \leq -1/2$? i get two positive/negative eigenvalues, but the first one is always zero, so i can't really say that they are maxima/minima, right? what's the method to determine if they are max/min? thank you very much

Last edited: Nov 16, 2013
2. Nov 16, 2013

### Felafel

Is it correct to say that, being semidefinite positive for $x \geq1/2$ the function is convex and has therefore infinite minima and being semidefinite negative for $x \leq -1/2$ it is concave and so all the points to the left of (-1/2, 0, 0) are maxima?