Răspuns:
[tex] \frac{\partial z}{\partial x}=3x^2-3=0, \; \frac{\partial z}{\partial y}=2y+2=0[/tex]. Rezulta [tex]x^2=1[/tex] deci [tex]x_1=1,x_2=-1[/tex]. Din 2y+2=0 rezulta y=-1.
Avem doua puncte critice: (1,-1), (-1,-1).
[tex] \frac{\partial^2 z}{\partial x^2}=6x, \frac{\partial^2 z}{\partial xy}=0, \frac{\partial^2 z}{\partial y^2}=2[/tex]
Matricea Hessiana este:
[tex]H_f(x,y)=\begin{pmatrix} \frac{\partial^2 z}{\partial x^2} & \frac{\partial^2 z}{\partial xy} \\ \frac{\partial^2 z}{\partial xy} &\frac{\partial^2 z}{\partial y^2}\end{pmatrix}=\begin{pmatrix} 6x & 0 \\ 0 & 2 \end{pmatrix} [/tex].
[tex]H_f(1,-1)=\begin{pmatrix} 6 & 0 \\ 0 & 2 \end{pmatrix} [/tex].
[tex]\Delta_1=6>0,\;\Delta_2=12>0[/tex] deci (1,-1) este punct de minim local.
[tex]H_f(-1,-1)=\begin{pmatrix} -6 & 0 \\ 0 & 2\end{pmatrix}[/tex].
[tex]\Delta_1=-6<0,\;\Delta_2=-12<0[/tex] deci (-1,-1) nu este punct de extrem.
