Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 14: Partial Derivatives - Section 14.7 - Extreme Values and saddle Points - Exercises 14.7 - Page 843: 42


Local Minimum: $f(\dfrac{1}{2}, 2 )=2-\ln \dfrac{1}{2}=2 +\ln 2$

Work Step by Step

$$f_x(x,y)=y+2-\dfrac{2}{x} \\ f_y(x,y)=x-\dfrac{1}{y}=0$$ The critical points are: $(\dfrac{1}{2}, 2 )$ Apply the second derivative test for the critical point $(\dfrac{1}{2}, 2 )$. $D(\dfrac{1}{2}, 2 )=f_{xx}f_{yy}-f^2_{xy}=(8)(\dfrac{1}{4})-1=2-1=1 \gt 0$ Now, we have the Local Minimum: $f(\dfrac{1}{2}, 2 )=2-\ln \dfrac{1}{2}=2 +\ln 2$
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