## Calculus 8th Edition

Minimum value: $f(1,\dfrac{1}{2})=-1$ Saddle point $(0,0)$
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. For $(x,y)=(1,\dfrac{1}{2})$ $D=108 \gt 0$ ; and $f_{xx} =36\gt 0$ Thus, when $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\gt 0$ , then $f(p,q)$ is a local minimum. For $(x,y)=(0,0)$ $D=-36 \lt 0$ 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, $f(1,\dfrac{1}{2})=x^3-6xy+8y^3=(1)^3-6(1)(1/2)+8(1/2)^3$ Therefore, we have Minimum value: $f(1,\dfrac{1}{2})=-1$ Saddle point $(0,0)$