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: 6


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

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 $\nabla f=\lt e^y,xe^y \gt$ and $\lambda \nabla g=\lambda \lt 2x,2y \gt$ From the given constraint condition $x^2+y^2=2$ we get, $e^y=\lambda 2x, xe^y=\lambda 2y$ Simplify. Thus, $x=\pm 1$ Since, $g(x,y)=x^2+y^2=2$ $\implies$ $y=-2,1$ When, $y=-2$, thus $x^2+y^2=2 \implies x^2=-2$; which is incorrect point. Hence, Maximum value is $f(1,1)=e$, Minimum value is $f(-1,1)=-e$
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