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: 12


Maximum:$1$ and minimum: $\dfrac{1}{3}$ or, Maximum:$f(0,0, \pm 1)= 1$ and minimum: $f(\pm \sqrt {\dfrac{1}{3}},\pm \sqrt {\dfrac{1}{3}},\pm \sqrt {\dfrac{1}{3}})=\dfrac{1}{3}$

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 $\nabla f=\lt 4x^3,4y^3,4z^3 \gt$ and $\lambda \nabla g=\lambda \lt 2x,2y,2z\gt$ From the given constraint condition $x^2+y^2+z^2=1$ we get, $4x^3=\lambda 2x ,4y^3=\lambda 2y,4z^3=\lambda 2z$ Simplify. Thus, $x^2=y^2=z^2$ Since, $g(x,y)=x^2+y^2+z^2=1 \implies$ $x=\pm \dfrac{1}{\sqrt 3}$ Now, we find that $x=y=z=\pm \sqrt {\dfrac{1}{3}}$ that are having eight different points with minimum value $\dfrac{1}{3}$ Also, we have possible points:$x=y=0,z=\pm 1$ that ar having two different points, with each of having maximum value :$1$ Hence, Maximum:$1$ and minimum: $\dfrac{1}{3}$ or, Maximum value is$f(0,0, \pm 1)= 1$ and minimum value is $f(\pm \sqrt {\dfrac{1}{3}},\pm \sqrt {\dfrac{1}{3}},\pm \sqrt {\dfrac{1}{3}})=\dfrac{1}{3}$
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