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: 11


The critical point is: $(\frac{4}{5},\frac{2}{5})$. There is a saddle point at $(\frac{4}{5},\frac{2}{5})$

Work Step by Step

$f(x,y)=x^{2}+xy-2x-2y+2$ $f_x(x,y)=2x+y-2=0 $ $f_y(x,y)=x-2y=0 \rightarrow x=2y$ Use the substitution method to solve the system of equations $f_x(x,y)=2x+y-2=0 \rightarrow 2(2y)+y-2=0$ $y=\frac{2}{5} \rightarrow x=\frac{4}{5}$ The critical point is:$(\frac{4}{5},\frac{2}{5})$ To identify any extrema: $f_{xx}(x,y)=2$ $f_{yy}(x,y)=-2$ $f_{xy}(x,y)=1$ For $(\frac{4}{5},\frac{2}{5})$: $f_{xx}(\frac{4}{5},\frac{2}{5})=2$ $f_{yy}(\frac{4}{5},\frac{2}{5})=-2$ $f_{xy}(\frac{4}{5},\frac{2}{5})=1$ $D=2\times(-2) - (1)^{2}=-5 \lt 0 \rightarrow$ There is a saddle point at $(\frac{4}{5},\frac{2}{5})$
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