Assume the function has a maximum. Find it. $f(x,y,z)=3-x^2-2y^2-3z^2-2xy-2xz$

Question

Putting the first derivatives equal to 0 I get a critical point at C(0,0,0), and then using the second derivatives I get a Determinant of the Hessian matrix=-24. Is C the maximum?

I have a doubt about the determinant. Should I put it in absolute values and look at f11,f22,f33 (which in this case are respectively -2,-4 and -6, so <0) or because it is Det=-24 the point is a saddle point? I am confused because our teacher told us to consider always the determinant in absolute values but in some books I found that when det<0 then the critical point is a saddle point. Thank you

Answer 1

The Hessian of $f$ is $$H(x,y,z)=-2\begin{bmatrix} 1 & 1 & 1\\ 1 & 2 & 0 \\ 1 & 0 & 3 \end{bmatrix} .$$ Since ALL the principal minor matrices have positive determinants (they are $1,1,1$), by the Sylvester's criterion and the negative factor $-2$, $H(0,0,0)$ is negative-definite and we may conclude that $(0,0,0)$ is a local maximum point.

The point $(0,0,0)$ is indeed a global maximum because: $$ \begin{align}f(x,y,z)&=3-\underbrace{\Big(\frac{x}{\sqrt{2}}+\sqrt{2}y\Big)^2}_{\geq 0}-\underbrace{\Big(\frac{x}{\sqrt{2}}+\sqrt{2}z\Big)^2}_{\geq 0}-\underbrace{\Big(z\Big)^2}_{\geq 0}\\ &\leq 3=f(0,0,0). \end{align}$$

Answer 2

The Hessian is$$\begin{bmatrix}-2 & -2 & -2 \\ -2 & -4 & 0 \\ -2 & 0 & -6\end{bmatrix},$$whose determinant is $-8$. More generally, its characteristic polynomial is$$P(\lambda)=-\lambda ^3-12 \lambda ^2-36 \lambda -8.$$Clearly, if $\lambda\geqslant0$, $P(\lambda)<0$. Therefore, all three roots of $P(\lambda)$ are all negative. It follows that indeed $f$ has a maximum at $(0,0,0)$.