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: 7


Maximum:$f(2,2,1)=9$, Minimum: $f(-2-2,-1)=-9$

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 $\nabla f=\lt 2,2,1 \gt$ and $\lambda \nabla g=\lambda \lt 2x,2y,2z\gt$ From the given constraint condition $x^2+y^2+z^2=9$ we get, $2=\lambda 2x, 2=\lambda 2y,1=\lambda 2z$ Simplify. Thus, $x=y=\pm 2$ Since, $g(x,y)=x^2+y^2+z^2=9$ $\implies$ $z=\pm 1$ Hence, Maximum value is $f(2,2,1)=9$, Minimum value is $f(-2-2,-1)=-9$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.