## Calculus 8th Edition

Maximum value: $f(1,1)=1$ Saddle point $(0,0),(0,3),(3,0)$
Second derivative test: Some noteworthy points to calculate the local minimum, local maximum and saddle point of $f$. 1. If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\gt 0$ , then $f(p,q)$ is a local minimum. 2.If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\lt 0$ , then $f(p,q)$ is a local maximum. 3. If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \lt 0$ , then $f(p,q)$ is a not a local minimum and local maximum or, a saddle point. Critical point are: $(1,1),(0,0),(0,3),(3,0)$ For $(x,y)=(1,1)$ $D=3 \gt 0$ ; and $f_{xx} =-2\gt 0$ Thus, when $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \lt 0$ and $f_{xx}(p,q)\gt 0$ , then $f(p,q)$ is a local maximum. For $(x,y)=(0,0)$ $D=-9 \lt 0$ For $(x,y)=(0,3)$ $D=-9 \lt 0$ For $(x,y)=(3,0)$ $D=-9 \lt 0$ When$D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \lt 0$ , then $f(p,q)$ is a not a local minimum and local maximum or, a saddle point. Now, $f(1,1)=3xy-x^2y-xy^2=3(1)(1)-(1)^2(1)-(1)(1)^2=1$ Therefore, we have Maximum value: $f(1,1)=1$ Saddle point $(0,0),(0,3),(3,0)$