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: 9


The critical point is $(\frac{2}{3},\frac{4}{3})$. There is a relative maximum at $(\frac{2}{3},\frac{4}{3})$

Work Step by Step

$f(x,y)=4xy-10x^{2}-4y^{2}+8x+8y+9$ $f_x(x,y)=4y-20x+8=0 \rightarrow x=\frac{2y+4}{10}$ $f_y(x,y)=4x-8y+8=0$ Use the substitution method to solve the system of equations $f_y(x,y)=4x-8y+8=0 \rightarrow 4(\frac{2y+4}{10})-8y+8=0$ $4y+8-40y+40=0 $ $y=\frac{4}{3}$ The critical point is: $(\frac{2}{3},\frac{4}{3})$ To identify any extrema: $f_{xx}(x,y)=-20$ $f_{yy}(x,y)=-8$ $f_{xy}(x,y)=4$ For $(\frac{2}{3},\frac{4}{3})$: $f_{xx}(\frac{2}{3},\frac{4}{3})=-20$ $f_{yy}(\frac{2}{3},\frac{4}{3})=-8$ $f_{xy}(\frac{2}{3},\frac{4}{3})=4$ $D=(-20)\times(-8) - (4)^{2}=144\gt 0$ and $f_{xx}(\frac{2}{3},\frac{4}{3})=-20 \lt0 \rightarrow$ There is a relative maximum at $(\frac{2}{3},\frac{4}{3})$
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