[tex]z=x^2+2y^2+6x-4y+22\\
z_x'=2x+6\\
z_y'=4y-4\\\\
2x+6=0\\
4y-4=0\\\\
2x=-6\\
4y=4\\\\
x=-3\\
y=1\\\\
[/tex]
We have a stationary point [tex](-3,1)[/tex]
[tex]z_{xx}''=2\\
z_{xy}''=z_{yx}=0\\
z_{yy}''=4[/tex]
[tex]2\cdot4-0^2=8 \ \textgreater \ 0[/tex] therefore there exist an extremum at this point.
[tex]z_{xx}''\ \textgreater \ 0[/tex] therefore the extremum indeed is the minimum.
[tex]z(-3,1)=(-3)^2+2\cdot1^2+6\cdot(-3)-4\cdot1+22=9+2-18-4+22\\
z(-3,1)=11[/tex]
So, the minimum is equal to 11.
