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: 4


There is a relative minimum at $(4,-2)$

Work Step by Step

$f_{x}(x,y)=2x + y - 6 = 0$ $f_{y}(x,y)=x + 2y=0 \rightarrow x=-2y$ Use the substitution method to solve the system of equations $2(-2y)+y-6=0 \rightarrow y=-2$ $x=4$ The critical points is $(4,-2)$ $f_{xx}(x,y)=2$ $f_{yy}(x,y)=2$ $f_{xy}(x,y)=1$ $f_{xx}(4,-2)=2$ $f_{yy}(4,-2)=2$ $f_{xy}(4,-2)=1$ $D=2*2-(1)^{2}=3$ Since there is D>0, and $f_{xx}(x,y)=2 \gt 0$, there is a relative minimum at $(4,-2)$
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