## Thomas' Calculus 13th Edition

Two Saddle points at $(0,\sqrt 5)$ and $(0,-\sqrt 5)$ ; Local minimum point at $f(2,1)=-30$ and Local maximum point at $f(-2,-1)=30$
Given: $f_x(x,y)=3x^2+3y^2-15=0, f_y(x,y)=6xy+3y^2-15=0$ Simplify the given two equations. Critical points: $(2,1),(-2,-1),(0,\sqrt 5), (0,-\sqrt 5)$ 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. $D(2,1)=f_{xx}f_{yy}-f^2_{xy}=180$ and $D=180 \gt 0$ and $f_{xx} \gt 0$ Thus, Local minimum point at $f(2,1)=-30$ For point $(-2,-1)$: $D=f_{xx}f_{yy}-f^2_{xy}=180$ and $D=180 \gt 0$ and $f_{xx} \lt 0$ so, Local maximum point at $f(-2,-1)=30$ For point $(0,\sqrt 5)$: $D=f_{xx}f_{yy}-f^2_{xy}=-180 \implies D=-180 \lt 0$ so, Saddle points at $(0,\sqrt 5)$ Local maximum point at $f(-2,0)=-4$ For point $(0,-\sqrt 5)$: $D=f_{xx}f_{yy}-f^2_{xy}=-180 \lt 0$ So, Saddle points at $(0,-\sqrt 5)$ Therefore, Two Saddle points at $(0,\sqrt 5)$ and $(0,-\sqrt 5)$ ; Local minimum point at $f(2,1)=-30$ and Local maximum point at $f(-2,-1)=30$