## Thomas' Calculus 13th Edition

Two Saddle points at $(0,1)$ and $(3,-2)$ ; Local minimum point at $f(3,1)=-34$ and Local maximum point at $f(0,-2)=20$
Given: $f_x(x,y)=6x^2-18x=0, f_y(x,y)=6y^2+6y-12=0$ Simplify the given two equations. Critical points: $(0,-2),(0,1),(3,-2), (3,1)$ 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. $D(0,-2)=f_{xx}f_{yy}-f^2_{xy}=180 \gt 0$ ; $f_{xx} \gt 0$ So, Local minimum point at $f(2,1)=-30$ For point $(-2,-1)$ : $D=f_{xx}f_{yy}-f^2_{xy}=180 \implies D=180 \gt 0$ and $f_{xx} \lt 0$ So, Local maximum point at $f(-2,-1)=30$ For point $(0,\sqrt 5)$: $D=f_{xx}f_{yy}-f^2_{xy}=-180 \implies D=-180 \lt 0$ Thus, Saddle points at $(0,1)$ $D(3,-2)=f_{xx}f_{yy}-f^2_{xy}=-180\implies D=-180 \lt 0$ Therefore, Two Saddle points at $(0,1)$ and $(3,-2)$ ; Local minimum point at $f(3,1)=-34$ and Local maximum point at $f(0,-2)=20$