Answer
Local maximum:$f(-2,-2)=4$
Work Step by Step
Given: $f(x,y)=xy-2x-2y-x^2-y^2$
This yields $f_x(x,y)=y-2-2x$ and $f_y(x,y)=x-2-2y$ gives $(x,y)=(-2,-2)$.
To solve this problem we will take the help of Second derivative test that suggests the following conditions to determine the local minimum, local maximum and saddle points of $f(x,y)$ or $f(x,y,z)$.
i) If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\gt 0$ , then $f(p,q)$ is a local minimum.
ii) If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\lt 0$ , then $f(p,q)$ is a local maximum.
iii) If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \lt 0$ , then $f(p,q)$ is a not a local minimum and local maximum or, a saddle point.
For $(x,y)=(-2,-2)$ , we have $D(-2,-2)=3 \gt 0$ and $f_{xx}(-2,-2)=2 \lt 0$
Hence, from the above conditions we conclude that
Local maximum:$f(-2,-2)=4$
