Use the principal minor criteria to determine (if possible) the nature of the critical points of the following functions :
1) $f(x ,y )=x^3+y^3-3x-12y+20$
2) $f(x,y,z)= 3x^2+2y^2+2z^2+2xy+2yz+2xz$
For the second question I can write
$$\begin{pmatrix} x & y & z \end{pmatrix} \begin{pmatrix} 3 & 1& 1\\ 1& 2 & 1\\ 1& 1 & 2 \end{pmatrix} \begin{pmatrix} x \\ y\\ z \end{pmatrix}=f(x,y,z)$$
That is $$x^T \cdot Ax=f(x,y,z)$$
Since $\Delta_1=3>0 , \Delta _2=5>0 $ and $\Delta _3=7>0$ , $A$ is positive definite.
Ho can I write $A$ for first question ? Any help ?
I will map it out and you can finish filling in the details. For the first problem, we have:
$$f(x ,y )=x^3+y^3-3x-12y+20$$
The critical points are given by:
$$\dfrac{\partial{f}}{\partial{x}} = 3x^2 - 3 = 0 \implies x = \pm 1 \\ \dfrac{\partial{f}}{\partial{y}} = 3y^2 - 12 = 0 \implies y = \pm 2$$
Thus, we have four critical points $(-1, -2), (-1, 2), (1, -2), (1, 2)$.
The Hessian matrix is given by:
$$H f(x, y) = \begin{pmatrix} 6x & 0\\ 0 & 6y \end{pmatrix}$$
The principal minors are given by:
$$\Delta_1= 6x, \Delta _2 = 36 x y$$
Now, when is the Hessian positive definite, negative definite and indefinite? These will tell you the local nature of those four critical points.
Even though this problem does not seem to ask for it, we can check for the global minima and maxima by finding:
$$\lim f(x, 0) ~\mbox{and}~ \lim f(0, y)$$
For this last part, you will find that there are no global minima or maxima.