## Thomas' Calculus 13th Edition

Local minimum of $f(0,0)=0$ and Saddle point at $(0,2)$.
Given: $f_x(x,y)=2x e^ {-y}=0, f_y(x,y)=2ye^{-y}-e^{-y}(x^2+y^2)=0$ Simplify the given two equations. Critical points: $(0, 0)$ and $(0,2)$ 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. $D(0,0)=f_{xx}f_{yy}-f^2_{xy}=4 \implies D=4 \gt 0;f_{xx}=2 \gt 0$ So, Local minimum of $f(0,0)=0$ $D(0,2)=f_{xx}f_{yy}-f^2_{xy}=-\dfrac{4}{e^4} \implies D=-\dfrac{4}{e^4} \lt 0$ So, Saddle point at $(0,2)$ Hence, Local minimum of $f(0,0)=0$ and Saddle point at $(0,2)$