## Thomas' Calculus 13th Edition

Local minimum of $f(-2,-2)=-8$
$f_x(x,y)=2x-y+2=0$ and $f_y(x,y)=-x+2y+2=0$ Simplify the above two equations to find the critical points. Thus, we have: $(-2,-2)$ By the second derivative test, for $(-2,-2)$ $D=f_{xx}f_{yy}-f^2_{xy}=(2)(2)-(1)^2=3 \lt 0$ Local minimum of $f(-2,-2)=-8$