Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 14 - Review - Exercises - Page 984: 63

Answer

$(\pm 3^{-1/4},3^{-1/4}\sqrt 2,\pm 3^{1/4}),(\pm 3^{-1/4},-3^{-1/4}\sqrt 2,\pm 3^{1/4})$

Work Step by Step

We need to apply the Lagrange Multipliers Method to determine the dimensions of a rectangular box of maximum volume. We have: $\nabla f=\lambda \nabla g$ The volume of a box is: $V=xyz$ or, $f=V=xyz$ Consider $\nabla f=\lt 2x,2y,2z \gt$ and $\lambda \nabla g=\lambda \lt y^2z^3,2xyz^3,3xy^2z^2 \gt$ Using the constraint condition we get, $yz=\lambda 2x, xz=\lambda 2y,xy=\lambda 2z$ Simplify to get the value of $x,y$ and $z$ We have $2x^2=y^2,3y^2=2z^2,xz=1$ From the given question, let us consider $g(x,y,z)=xy^2z^3$ yields $x=\pm \dfrac{1}{3^{1/4}}$ Thus, $y=\pm \dfrac{\sqrt 2}{3^{1/4}}$ and $z=\pm 3^{1/4}$ Hence, the points which are closet to the origin are: $(\pm 3^{-1/4},3^{-1/4}\sqrt 2,\pm 3^{1/4}),(\pm 3^{-1/4},-3^{-1/4}\sqrt 2,\pm 3^{1/4})$
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