Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 14 - Section 14.8 - Lagrange Multipliers - 14.8 Exercise - Page 977: 3


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

Work Step by Step

Use Lagrange Multipliers Method: $\nabla f(x,y)=\lambda \nabla g(x,y)$ This yields $\nabla f=\lt x,-y \gt$ and $\lambda \nabla g=\lambda \lt x,y \gt$ Using the constraint condition $x^2+y^2=1$ we get, $x=\lambda x, -y=\lambda y$ After solving, we get $x=0$ and $\lambda =1$ Since, $g(x,y)=x^2+y^2=1$ yields $y=\pm 1$ When $\lambda =1$, then $x=\pm 1, y=0$ Hence, Maximum:$f(\pm 1,0)=1$, Minimum: $f(0, \pm 1)=-1$
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