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: 12


The critical point is: $(2,-1)$. There is a relative maximum at $(2,-1)$

Work Step by Step

$f(x,y)=x^{2}+xy+y^{2}-3x-5$ $f_x(x,y)=2x+y-3=0 $ $f_y(x,y)=x+2y=0 \rightarrow x=-2y$ Use the substitution method to solve the system of equations $f_x(x,y)=2x+y-3=0 \rightarrow 2(-2y)+y-3=0$ $y=-1 \rightarrow x=2$ The critical point is: $(2,-1)$ To identify any extrema: $f_{xx}(x,y)=2$ $f_{yy}(x,y)=2$ $f_{xy}(x,y)=1$ For (2,-1): $f_{xx}(2,-1)=2$ $f_{yy}(2,-1)=2$ $f_{xy}(2,-1)=1$ $D=2\times2 - (1)^{2}=3 \gt0$ and $f_{xx}(2,-1)=2\gt0 \rightarrow$ There is a relative maximum at $(2,-1)$
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