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: 21

Answer

Maximum: $f(\dfrac{3}{\sqrt 2},- \dfrac{3}{\sqrt 2})=9+12 \sqrt2 $ and Minimum: $f(-2,2)=-8$

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 $\nabla f(x,y)=\lt 2x+4,2y-4 \gt$ and $\lambda g(x,y)= \lt 2x,2y \gt$ Using the constraint condition $x^2+y^2 \leq 9$ we get, $x^2=\pm \dfrac{9}{2}$ and $x= \pm \dfrac{3}{\sqrt 2}$ $\nabla f(x,y)=\lt 2x+4,2y-4 \gt$ and $\lambda g(x,y)= \lt 2x,2y \gt$ we get $y=-x$ and $y=-\dfrac{3}{\sqrt 2}, \dfrac{3}{\sqrt 2}$ Hence, Maximum: $f(\dfrac{3}{\sqrt 2},- \dfrac{3}{\sqrt 2})=9+12 \sqrt2 $ and Minimum: $f(-2,2)=-8$
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