Calculus 8th Edition

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

Chapter 14 - Partial Derivatives - 14.8 Lagrange Multipliers - 14.8 Exercises - Page 1017: 3


Maximum:$f(\pm 1,0)=1$, Minimum: $f(0, \pm 1)=-1$

Work Step by Step

Our aim is to calculate the extreme values with the help of Lagrange Multipliers Method subject to the given constraints. For this, we have:$\nabla f(x,y)=\lambda \nabla g(x,y)$ This yields $\nabla f(x,y)=\lt x,-y \gt$ and $\lambda \nabla g(x,y)=\lambda \lt x,y \gt$ From the given question, by using the constraint condition $x^2+y^2=1$ we get, $x=\lambda x, -y=\lambda y$ After simplifications, we get $x=0$ and $\lambda =1$ As we are given that $g(x,y)=x^2+y^2=1$ yields $y=\pm 1$ When $\lambda =1$, then $x=\pm 1, y=0$ Therefore, Maximum value is $f(\pm 1,0)=1$, Minimum value is $f(0, \pm 1)=-1$
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