Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 14 - Section 14.8 - Lagrange Multipliers - 14.8 Exercise - Page 1027: 23

Answer

Minimum: $f(1,1)=f(-1,-1)=2$

Work Step by Step

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