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: 1


Local minimum of $f(-3,3)=-5$

Work Step by Step

Since, we have $f_x(x,y)=2x+y+3=0, f_y(x,y)=x+2y-3=0$ After solving the above two equations, we get $x=-3,y=3$ Thus the critical point is: $(-3,3)$ As per second derivative test, we have $D=f_{xx}f_{yy}-f^2_{xy}=3$ and $D=3 \gt 0$ and $f_{xx}(-3,3)=2 \gt 0$ Thus, local minimum of $f(-3,3)=-5$
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