Question

Please help me!!!! What is the minimum value of z if z=x^2+2y^2+6x-4y+22?

Answer 1

`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\\\\`

We have a stationary point
`(-3,1)`


`z_(xx)''=2\\ z_(xy)''=z_(yx)=0\\ z_(yy)''=4`


`2\cdot4-0^2=8 \ \textgreater \ 0` therefore there exist an extremum at this point.


`z_(xx)''\ \textgreater \ 0` therefore the extremum indeed is the minimum.


`z(-3,1)=(-3)^2+2\cdot1^2+6\cdot(-3)-4\cdot1+22=9+2-18-4+22\\ z(-3,1)=11`

So, the minimum is equal to 11.