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: 20


Maximum: $f(2,1,0)=5$ and Minimum: $f(0,-1,0)=1$

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 in 3D as: $\nabla f(x,y,z)=\lambda_1 \nabla g(x,y,z)+\lambda_2 \nabla h(x,y,z)$ This yields $\nabla f=\lt 2x,2y,2z \gt$ and $\lambda_1 \nabla g(x,y,z)+\lambda_2 \nabla h(x,y,z)= \lt \lambda_1,-\lambda_1+2\lambda_2y,-2\lambda_2 z \gt$ As per the given constraint condition $x-y=1$we get, $x=2/3 ,y=-1/3$ We can see that when we use the values $x=2/3 ,y=-1/3$ of $x,y$ in another constraint $y^2-z^2=1$, the invalid value will be achieved. In order to get the valid or value need to consider $z=0$; after simplifications, we have $x=2,0$ and $y=1,-1$ Thus, Maximum value is $f(2,1,0)=5$ and Minimum value is $f(0,-1,0)=1$
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