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: 6


There is a relative minimum at $(\frac{-25}{3}, 5)$

Work Step by Step

We are given $f(x,y)=2x^{2}+3xy+2y^{2}-5x+5y$ $f_{x}(x,y)=4x +3y -5 = 0$ $f_{y}(x,y)=3x + 4y+5= 0 $ $\rightarrow x=\frac{-4y-5}{3}$ Use the substitution method to solve the system of equations $4(\frac{-4y-5}{3}) +3y -5 =0 $ $-16y-20+9y-15=0 \rightarrow y=5$ $x=\frac{-25}{3}$ The critical points is $(\frac{-25}{3}, 5)$ $f_{xx}(x,y)=4$ $f_{yy}(x,y)=4$ $f_{xy}(x,y)=3$ $f_{xx}(\frac{-25}{3}, 5)=4$ $f_{yy}(\frac{-25}{3}, 5)=4$ $f_{xy}(\frac{-25}{3}, 5)=3$ $D=4*4-(3)^{2}=7$ Since D>0 and $f_{xx}(\frac{-25}{3}, 5)=4 \gt 0$, there is a relative minimum at $(\frac{-25}{3}, 5)$
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