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: 31


$\dfrac{2}{\sqrt 3}$ or, $\dfrac{2\sqrt 3}{3}$

Work Step by Step

Given: Equation of plane: $x+y+z=1$ and point $(2,0,-3)$ Need to apply Lagrange Multipliers Method to determine the shortest distance from the point $(2,0,-3)$ to the plane $x+y+z=1$. We have $\nabla f=\lambda \nabla g$ Use the formula to calculate the shortest distance. $D=|\dfrac{pa+qb+rc-s}{\sqrt{p^2+q^2+r^2}}|=|\dfrac{(1)(2)+(1)(0)+(1)(-3)-1}{\sqrt{1^2+1^2+1^2}}|$ Hence, $D=\dfrac{2}{\sqrt 3}$ or, $\dfrac{2\sqrt 3}{3}$
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