Calculus 8th Edition

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

Chapter 14 - Partial Derivatives - Review - Exercises - Page 1024: 59


Maximum value is $\frac{2}{3\sqrt 3}$ Minimum value is $-\frac{2}{3\sqrt 3}$

Work Step by Step

$f(x,y)=x^2y$; $x^2+y^2=1$ According to Lagrange multipliers, we have the following equations: $2xy= \lambda (2x)$ ...(1) $x^2= \lambda (2y)$... (2) Divide equation (1) by (2), we get $2y^2=x^2$ Thus, $2y^2+y^2=1$ $y= \pm \frac{1}{\sqrt 3}$ which means $x^2=\frac{2}{3}$ Therefore, Maximum value is $\frac{2}{3}\times\frac{1}{\sqrt 3}=\frac{2}{3\sqrt 3}$ Minimum value is $\frac{2}{3}\times\frac{-1}{\sqrt 3}=-\frac{2}{3\sqrt 3}$
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