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: 5


There is a relative minimum at $(-2,-2)$

Work Step by Step

$f_{x}(x,y)=2x - y + 2 = 0$ $f_{y}(x,y)=-x + 2y+2= 0 \rightarrow x=2y+2$ Use the substitution method to solve the system of equations $2(2y+2) - y + 2 =0 \rightarrow y=-2$ $x=-2$ The critical point is $(-2,-2)$ $f_{xx}(x,y)=2$ $f_{yy}(x,y)=2$ $f_{xy}(x,y)=-1$ $f_{xx}(-2,-2)=2$ $f_{yy}(-2,-2)=2$ $f_{xy}(-2,-2)=-1$ $D=2*2-(-1)^{2}=3$ Since there is D>0, and $f_{xx}(x,y)=2 \gt 0$, there is a relative minimum at $(-2,-2)$
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