Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 14 - Partial Derivatives - Review - Exercises - Page 1024: 60


Maximum value is $\sqrt 2$ Minimum value is $-\sqrt 2$

Work Step by Step

$f(x,y)=\frac{1}{x}+\frac{1}{y}$; $\frac{1}{x^2}+\frac{1}{y^2}$ According to Lagrange multipliers, we have the following equations: $\frac{1}{x^2}= \lambda (\frac{-2}{x^3})$ ...(1) $\frac{1}{y^2}= \lambda (\frac{-2}{y^3})$... (2) From equations (1) by (2), we get $x=y$ Thus, $x^2=y^2=2$ Therefore, Maximum value is $\frac{1}{\sqrt 2}+\frac{1}{\sqrt 2}=\sqrt 2$ Minimum value is $-\frac{1}{\sqrt 2}-\frac{1}{\sqrt 2}=-\sqrt 2$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.