Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 14 - Partial Derivatives - 14.7 Exercises - Page 977: 3

Answer

$f(x,y)$ has a local minimum at $(1,1)$ and a saddle point at $(0,0)$.

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 not a local minimum, maximum or a saddle point. For the given question, we have $f_x(x,y)=3x^2-3y$ and $f_y(x,y)=3y^2-3x$ gives $(x,y)=(0,0)$ and $(1,1)$. For $(x,y)=(0,0)$ $D(0,0)=-9 \lt 0$ For $(x,y)=(1,1)$ $D(1,1)=27 \gt 0$ and $f_{xx}(1,1)=6 \gt 0$ This implies that $f(x,y)$ has a local minimum at $(1,1)$ and saddle point at $(0,0)$.
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