Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 14 - Section 14.8 - Lagrange Multipliers - 14.8 Exercise - Page 977: 9

Answer

Maximum:$f(1, \pm\sqrt2,1)=f(-1, \pm\sqrt2,-1)=2$, Minimum: $f(1, \pm\sqrt2,-1)=f(-1, \pm\sqrt2,1)=-2$

Work Step by Step

Use Lagrange Multipliers Method: $\nabla f(x,y,z)=\lambda \nabla g(x,y,z)$ This yields $\nabla f=\lt y^2z,2xyz,xy^2 \gt$ and $\lambda \nabla g=\lambda \lt 2x,2y,2z\gt$ Using the constraint condition $x^2+y^2+z^2=4$ we get, $y^2z=\lambda 2x, 2xyz=\lambda 2y,xy^2=\lambda 2z$ After solving, we get $x^2=z^2$ and $y^2=2z^2$ Since, $g(x,y)=x^2+y^2+z^2=4$ gives $z=\pm 1$ Thus, $x=\pm 1$ and $y=\pm \sqrt2$ Hence, Maximum:$f(1, \pm\sqrt2,1)=f(-1, \pm\sqrt2,-1)=2$, Minimum: $f(1, \pm\sqrt2,-1)=f(-1, \pm\sqrt2,1)=-2$
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