## Thomas' Calculus 13th Edition

Two Saddle points at $(0,0) , (-2,2)$; Local minimum at $(0,2)=-12$ and Local maximum at $(-2,0)=-4$
Given: $f_x(x,y)=3x^2+6x=0, f_y(x,y)=3y^2-6y=0$ Simplify the given two equations. Critical points: $(0,0),(0,2),(-2,0), (-2,2)$ In order to solve this problem we will have to apply Second derivative test that suggests the following conditions to calculate the local minimum, local maximum and saddle points of $f(x,y)$ or $f(x,y,z)$. 1. If $D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2 \gt 0$ and $f_{xx}(a,b)\gt 0$ , then $f(a,b)$ is a local minimum. 2. If $D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2 \gt 0$ and $f_{xx}(a,b)\lt 0$ , then $f(a,b)$ is a local maximum. 3. If $D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2 \lt 0$ , then $f(a,b)$ is a not a local minimum or local maximum but a saddle point. $D(0,0)=f_{xx}f_{yy}-f^2_{xy}=-36$ and $D=-36 \lt 0$ Thus, Saddle point at $(0,2)$ $D(0,2)=f_{xx}f_{yy}-f^2_{xy}=36 \gt 0$ and $f_{xx} \gt 0$ So, Local minimum point at $f(0,2)=-12$ $D(-2,0)=f_{xx}f_{yy}-f^2_{xy}=36$ and $D=36 \gt 0$ and $f_{xx} \lt 0$ Thus, Local maximum point at $f(-2,0)=-4$ $D(-2,2)=f_{xx}f_{yy}-f^2_{xy}=-36$ and $D=-36 \lt 0$ Hence, Two Saddle points at $(0,0) , (-2,2)$; Local minimum at $(0,2)=-12$ and Local maximum at $(-2,0)=-4$