## Thomas' Calculus 13th Edition

Saddle point at $(0,0)$; Local maximum at $(\dfrac{-2}{3},\dfrac{2}{3})=\dfrac{170}{27}$
Given: $f_x(x,y)=3x^2-2y=0, f_y(x,y)=-3y^2-2x=0$ Simplify the given two equations. Critical point: $(0,0)$ and $(\dfrac{-2}{3},\dfrac{2}{3})$ 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. $D(0,0)=f_{xx}f_{yy}-f^2_{xy}=-4$ and $D=-4 \lt 0$ So, Saddle point at $(0,0)$ $D(\dfrac{-2}{3},\dfrac{2}{3})=f_{xx}f_{yy}-f^2_{xy}=12$ and $D=12 \gt 0$ and $f_{xx}=-4 \lt 0$ Thus, Local maximum at $(\dfrac{-2}{3},\dfrac{2}{3})=\dfrac{170}{27}$