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: 15


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

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 $\nabla f(x,y)=\lt 2x,2y \gt$ and $\lambda \nabla g(x,y)=\lambda \lt y,x \gt$ As per the given constraint condition, we have $xy=1$ we get, $2x=\lambda y, 2y=\lambda x$ Simplify. Thus, $x=\pm 1$ Since, $g(x,y)=xy=1 \implies $$y=\pm 1$ We can see that the function has no maximum value, only possess minimum value $f(1,1)=f(-1,-1)=2$.
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