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 14.8 - Lagrange Multipliers - Exercises 14.8 - Page 852: 1


$(\pm \dfrac{1}{\sqrt 2},\dfrac{1}{2}) $ and $(\pm \dfrac{1}{\sqrt 2},-\dfrac{1}{2})$

Work Step by Step

Given: $f(x,y) =xy$ and $g(x,y)=x^2+2y^2-1=0$ ...(1) Consider the gradient equation as follows: $\nabla =\lambda \nabla g$ Here $y=2 \lambda x $ or, $y=\pm \dfrac{1}{2 \sqrt 2}$ and $x=4 \lambda y $ or, $x=\pm \dfrac{2}{2 \sqrt 2}y$ Now, equation (1) becomes: $g(x,y)=x^2+2y^2-1=0$ This implies that $x=\pm \dfrac{1}{\sqrt 2}$ and $y= \pm \dfrac{1}{2}$ Answer: our points are: $(\pm \dfrac{1}{\sqrt 2},\dfrac{1}{2}) $ and $(\pm \dfrac{1}{\sqrt 2},-\dfrac{1}{2})$
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.