Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 14 - Partial Derivatives - 14.7 Maximum and Minimum Values - 14.7 Exercises - Page 1009: 42



Work Step by Step

Given: Equation of plane: $x-2y+3z=6$ ; point $(0,1,1)$ Need to apply Lagrange Multipliers Method to determine the shortest distance from the point $(0,1,1)$ to the plane $x-2y+3z=6$ we have $\nabla f=\lambda \nabla g$ Since, we have the equation of plane $x-2y+3x=6$ which yields $0+(1)(p)-2(1-2(p))+3(1+3(p))=6$ Simplify. $p=\dfrac{5}{14}$ Therefore, $x=p,y=1-2p,z=1+3p$ it gives $x=\dfrac{5}{14}$, $y=\dfrac{2}{7}$ and $z=\dfrac{29}{14}$ Hence, our result is: $(\dfrac{5}{14},\dfrac{2}{7},\dfrac{29}{14})$
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