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: 7


The critical point is: (15,-8). There is a relative minimum at (15,-8)

Work Step by Step

$f(x,y)=x^{2}+3xy+3y^{2}-6x+3y$ $f_x(x,y)=2x+3y-6=0$ $f_y(x,y)=3x+6y+3=0 \rightarrow x=-2y-1$ Use the substitution method to solve the system of equations: $f_y(x,y)=2x+3y-6=0 \rightarrow 2(-2y-1)+3y-6=0$ $y=-8$ The critical point is: $(15,-8)$ To identify any extrema: $f_{xx}(x,y)=2$ $f_{yy}(x,y)=6$ $f_{xy}(x,y)=3$ For $(15,-8)$: $f_{xx}(15,-8)=2$ $f_{yy}(15,-8)=6$ $f_{xy}(15,-8)=3$ $D=2\times6 - (3)^{2}=3\gt 0$ and $f_{xx}(15,-8)=2\gt0 \rightarrow$ There is a relative minimum at $(15,-8)$
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