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: 13


Maximum:$f(\dfrac{1}{2},\dfrac{1}{2}, \dfrac{1}{2},\dfrac{1}{2}) =2$ and minimum: $f(-\dfrac{1}{2},-\dfrac{1}{2}, -\dfrac{1}{2},-\dfrac{1}{2}) =-2$

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 $\nabla f=\lt 1,1,1,1 \gt$ and $\lambda \nabla g=\lambda \lt 2x,2y,2z,2t\gt$ From the given constraint condition $x^2+y^2+z^2+t^2=1$ we get, $1=\lambda 2x ,1=\lambda 2y,1=\lambda 2z,1=\lambda 2t$ Simplify. Thus, $x=y=z=t$ Since, $g(x,y)=x^2+y^2+z^2+t^2=1 \implies$ $x=\pm \dfrac{1}{2}$ Also, $x=y=z=t= \pm \dfrac{1}{2}$ Hence, Maximum value is $f(\dfrac{1}{2},\dfrac{1}{2}, \dfrac{1}{2},\dfrac{1}{2}) =2$ and minimum value is $f(-\dfrac{1}{2},-\dfrac{1}{2}, -\dfrac{1}{2},-\dfrac{1}{2}) =-2$
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