Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 14 - Partial Derivatives - 14.7 Maximum and Minimum Values - 14.7 Exercises - Page 1008: 7

Answer

No local minimum and no local maximum. Saddle point at $(1,1),(-1,-1)$.

Work Step by Step

Given: $f(x,y)=(x-y)(1-xy)$ This yields $f_x(x,y)=1+y^2-2xy$ and $f_y(x,y)=x^2-1+2xy$ gives $(x,y)=(1,1),(-1,-1)$. To solve this problem we will take the help of Second derivative test that suggests the following conditions to determine the local minimum, local maximum and saddle points of $f(x,y)$ or $f(x,y,z)$. i) 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. ii) 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. iii) 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 $(x,y)=(1,1)$, we have $D(1,1)=-4 \lt 0$ and $D(-1,-1)=-4 \lt 0$ Hence, No local minimum and no local maximum. Saddle point at $(1,1),(-1,-1)$.
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