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: 11


Maximum:$\sqrt 3$ and minimum: $1$

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 $\nabla f=\lt 2x,2y,2z \gt$ and $\lambda \nabla g=\lambda \lt 4x^3,4y^3,4z^3\gt$ From the given constraint condition $x^4+y^4+z^4=1$ we get, $2x=\lambda 4x^3,2y=\lambda 4y^3,2z=\lambda 4z^3$ Simplify. Thus, $x=y=z$ Since, $g(x,y)=x^4+y^4+z^4=1 \implies$ $x=\sqrt[4] {\dfrac{1}{3}}$ Now, $x=y=z=\sqrt[4] {\dfrac{1}{3}}$ Also, we have possible points:$x=y=0,z=\pm 1$ Hence, Maximum value is $\sqrt 3$ and minimum value is $1$
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