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: 7

Answer

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

Work Step by Step

Use Lagrange Multipliers Method: $\nabla f(x,y,z)=\lambda \nabla g(x,y,z)$ This yields $\nabla f=\lt 2,2,1 \gt$ and $\lambda \nabla g=\lambda \lt 2x,2y,2z\gt$ Using the constraint condition $x^2+y^2+z^2=9$ we get, $2=\lambda 2x, 2=\lambda 2y,1=\lambda 2z$ After solving, we get $x=y=\pm 2$ Since, $g(x,y)=x^2+y^2+z^2=9$ gives $z=\pm 1$ Hence, Maximum:$f(2,2,1)=9$. Minimum: $f(-2-2,-1)=-9$
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