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: 19


Maximum: $\dfrac{3}{2}$ and Minimum: $\dfrac{1}{2}$

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:$\nabla f(x,y)=\lambda \nabla g(x,y)$ RE-write as: $\nabla f(x,y,z)=\lambda_1 \nabla g(x,y,z)+\lambda_2 \nabla h(x,y,z)$ This yields $\nabla f=\lt a,z+x,y, \gt$ and $\lambda_1 \nabla g(x,y,z)+\lambda_2 \nabla h(x,y,z)= \lt ay,ax+2by,2bz \gt$ As per the given constraint condition we get, $x=\pm \sqrt 2$ Simplify. Thus, $xy=1$ and $y^2+z^2=1$ we get $z=\pm \dfrac{1}{\sqrt2}$ and $y=1-z^2=1-\pm \dfrac{1}{\sqrt2}=\pm \dfrac{1}{\sqrt2}$ The desired results are: Maximum: $\dfrac{3}{2}$ and Minimum: $\dfrac{1}{2}$
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