University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 13 - Section 13.7 - Extreme Values and Saddle Points - Exercises - Page 737: 6

Answer

Saddle point at $(2,1)$

Work Step by Step

Since, we have $f_x(x,y)=2x-4y=0, f_y(x,y)=-4x+2y+6=0$ After solving the above two equations, we get $x=2,y=1$ Thus the critical point is: $(2,1)$ As per second derivative test, we have $D=f_{xx}f_{yy}-f^2_{xy}=-12$ and $D=-12 \lt 0$ Thus, saddle point at $(2,1)$
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