## Calculus 8th Edition

Published by Cengage

# Chapter 14 - Partial Derivatives - 14.7 Maximum and Minimum Values - 14.7 Exercises - Page 1008: 11

#### Answer

Maximum value: $f(-1,0) =2$, Minimum value: $f(1,0) =-2$ Saddle point:$(0, \pm 1)$

#### Work Step by Step

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 $(x,y)=( -1,0)$ $D(-1,0)=36 \gt 0$ ; and $f_{xx}(-1,0)=6 \gt 0$ For $(x,y)=(1,0)$ $D(1,0)=36 \gt 0$ ; and $f_{xx}(-1,0)=-6 \lt 0$ $D(0, \pm 1)=-36 \lt 0$; saddle points Hence, Maximum value: $f(-1,0) =2$, Minimum value: $f(1,0) =-2$ Saddle point:$(0, \pm 1)$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.