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: 4


Maximum:$f(3,1)=10$, Minimum: $f(-3,-1)=-10$

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 $\nabla f(x,y)=\lt 3,1 \gt$ and $\lambda \nabla g(x,y)=\lambda \lt 2x,2y \gt$ From the given question, by using the constraint condition $x^2+y^2=10$ we get, $3=\lambda 2x, 1=\lambda 2y$ After simplifications, we get $x=\pm 3$ Since, $g(x,y)=x^2+y^2=10$ $\implies$ $y=\pm 1$ Hence, Maximum value is $f(3,1)=10$, Minimum value is $f(-3,-1)=-10$
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