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: 5


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

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 $\nabla f=\lt y,x \gt$ and $\lambda \nabla g=\lambda \lt 8x,2y \gt$ From the given constraint condition $4x^2+y^2=8$ we get, $y=\lambda 8x, x=\lambda 2y$ After simplifications, we get $x=\pm 1$ Since, $g(x,y)=x^2+y^2=10$ $\implies$ $y=2$ Hence, Maximum value is $f(1,2)=f(-1,-2)=2$, Minimum value is $f(-1,2)=f(1,-2)=-2$
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