## Calculus with Applications (10th Edition)

There is a relative minimum at $(-2,-2)$
$f_{x}(x,y)=2x - y + 2 = 0$ $f_{y}(x,y)=-x + 2y+2= 0 \rightarrow x=2y+2$ Use the substitution method to solve the system of equations $2(2y+2) - y + 2 =0 \rightarrow y=-2$ $x=-2$ The critical point is $(-2,-2)$ $f_{xx}(x,y)=2$ $f_{yy}(x,y)=2$ $f_{xy}(x,y)=-1$ $f_{xx}(-2,-2)=2$ $f_{yy}(-2,-2)=2$ $f_{xy}(-2,-2)=-1$ $D=2*2-(-1)^{2}=3$ Since there is D>0, and $f_{xx}(x,y)=2 \gt 0$, there is a relative minimum at $(-2,-2)$