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 1018: 35


$(\dfrac{100}{3} ,\dfrac{100}{3} ,\dfrac{100}{3})$

Work Step by Step

Given: $x+y+z=100$ Re-write the given equations as: $z=100-x-y$ To solve this problem we will take the help of Second derivative test that suggests the following conditions to determine the local minimum, local maximum and saddle points of $f(x,y)$ or $f(x,y,z)$. i) If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\gt 0$ , then $f(p,q)$ is a local minimum. ii) If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\lt 0$ , then $f(p,q)$ is a local maximum. iii) If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \lt 0$ , then $f(p,q)$ is a not a local minimum and local maximum or, a saddle point. $D(\dfrac{100}{3},\dfrac{100}{3})=\dfrac{200}{3} \gt 0$ and $f_{xx}=-\dfrac{200}{3} \lt 0$ Hence, our result is: $(\dfrac{100}{3} ,\dfrac{100}{3} ,\dfrac{100}{3})$
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