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: 14


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

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:$\nabla f(x,y)=\lambda \nabla g(x,y)$ Re-write as: $\nabla f(x_1,x_2,...,x_n)=\lambda \nabla g(x_1,x_2,....,x_n)$ This yields $\nabla f=\lt 1,1,1,....,1 \gt$ and $\lambda \nabla g=\lambda \lt 2x_1,2x_2,2x_3,....,2x_n\gt$ As per given constraint condition, we have $x^2+y^2+z^2+t^2=1$ we get, $1=\lambda 2x_1 ,...,1=\lambda 2x_n$ Simplify. Thus, $x_1=x_2=x_3=...=N$ Now, $x_1=x_2=....= \pm \dfrac{1}{\sqrt n}$ We find that the Maximum value is $f(\dfrac{1}{\sqrt n},\dfrac{1}{\sqrt n}, ......,\dfrac{1}{\sqrt n}) =\sqrt n$ and minimum value is $f(-\dfrac{1}{\sqrt n},-\dfrac{1}{\sqrt n}, ......,-\dfrac{1}{\sqrt n}) =-\sqrt n$
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