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: 8


The critical point is $(-8,-23)$. There is a relative maximum at $(-8,-23)$

Work Step by Step

$f(x,y)=5xy-7x^{2}-y^{2}+3x-6y-4$ $f_x(x,y)=5y-14x+3=0 \rightarrow x=\frac{5y+3}{14}$ $f_y(x,y)=5x-2y-6=0$ Use the substitution method to solve the system of equations $f_y(x,y)=5x-2y-6=0 \rightarrow 5(\frac{5y+3}{14})-2y-6=0$ $25y+15-28y-84=0 $ $y=-23$ The critical point is: $(-8,-23)$ To identify any extrema: $f_{xx}(x,y)=-14$ $f_{yy}(x,y)=-2$ $f_{xy}(x,y)=5$ For $(-8,-23)$: $f_{xx}(-8,-23)=-14$ $f_{yy}(-8,-23)=-2$ $f_{xy}(-8,-23)=5$ $D=(-14)\times(-2) - (5)^{2}=3\gt 0$ and $f_{xx}(-8,-23)=-14 \lt0 \rightarrow$ There is a relative maximum at $(-8,-23)$
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