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Dheeraj Inampudi · Answer 1 · 2018-05-04T13:05:33+0000

$z=x^2+2y^2+6x-4y+22\\ z_x'=2x+6\\ z_y'=4y-4\\\\ 2x+6=0\\ 4y-4=0\\\\ 2x=-6\\ 4y=4\\\\ x=-3\\ y=1\\\\$

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.

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