Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 14 - Partial Derivatives - 14.8 Exercises - Page 987: 8

Answer

Minimum:$f(4,4,4)=48$. Maximum value does not exist.

Work Step by Step

Use Lagrange Multipliers Method: $\nabla f(x,y,z)=\lambda \nabla g(x,y,z)$ This yields $\nabla f=\lt 2x,2y,2z \gt$ and $\lambda \nabla g=\lambda \lt 1,1,1\gt$ Using the constraint condition we get, $2x=\lambda, 2y=\lambda,2z=\lambda $ After solving, we get $x=y=z$ Since, $g(x,y)=x+y+z=12$ gives $x=\lambda/12$ Thus, $\lambda=8$, so $x=y=z=4$ Hence, Minimum:$f(4,4,4)=48$. Maximum value does not exist.
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