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: 10


Maximum value: $f(1,0)=f(-1,0) =3$ Saddle point:$(0,0)$

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 $(x,y)=( 1,0)$ $D(1,0)=16 \gt 0$ ; and $f_{xx}(1,0)=-8 \gt 0$ For $(x,y)=( -1,0)$ $D(-1,0)=16 \gt 0$ ; and $f_{xx}(-1,0)=-8 \gt 0$ $D(0,0)=-8 \lt 0$; saddle points As we are given that $f(x,y)=2-x^4+2x^2-y^2$ $f(1,0)=2-(1)^4+2(1)^2-(0)^2=3$ $f(-1,0)=2-(-1)^4+2(-1)^2-(0)^2=3$ Hence, Maximum value: $f(1,0)=f(-1,0) =3$ Saddle point:$(0,0)$
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