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: 8


Maximum:$f(2,2\sqrt2,2 \sqrt2)=e^{16}$, Minimum: $f(-2,-2\sqrt2,-2 \sqrt2)=e^{-16}$ or, Maximum:$f(2,\sqrt8, \sqrt 8)=e^{16}$, Minimum: $f(-2,-\sqrt 8,-\sqrt 8)=e^{-16}$

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 $\nabla f=\lt yze^{xyz},xze^{xyz},xye^{xyz} \gt$ and $\lambda \nabla g=\lambda \lt 4x,2y,2z\gt$ From the given constraint condition $2x^2+y^2+z^2=24$ we get, $yze^{xyz}=\lambda 4x, xze^{xyz}=\lambda 2y,xye^{xyz}=\lambda 2z$ Simplify. Thus, $y=z=2x^2$ Since, $g(x,y)=2x^2+y^2+z^2=24$ $\implies$ $x=\pm 2$ and $y=z=\pm 2\sqrt2$ Hence, Maximum va;ue is $f(2,2\sqrt2,2 \sqrt2)=e^{16}$, Minimum value is $f(-2,-2\sqrt2,-2 \sqrt2)=e^{-16}$ or, Maximum value is $f(2,\sqrt8, \sqrt 8)=e^{16}$, Minimum value is $f(-2,-\sqrt 8,-\sqrt 8)=e^{-16}$
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