Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 9 - Multiveriable Calculus - 9.3 Maxima and Minima - 9.3 Exercises - Page 488: 2


There is a saddle point at $(-2,\frac{5}{3})$

Work Step by Step

$f_{x}(x,y)=3y -5 = 0 \rightarrow y=\frac{5}{3}$ $f_{y}(x,y)=3x+6=0 \rightarrow x=-2$ The critical points is $(-2,\frac{5}{3})$ $f_{xx}(x,y)=0$ $f_{yy}(x,y)=0$ $f_{xy}(x,y)=3$ $D=0.0-3^{2}=-9$ Since there is D<0, there is a saddle point at $(-2,\frac{5}{3})$
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