Calculus: Early Transcendentals 8th Edition

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

Chapter 14 - Section 14.8 - Lagrange Multipliers - 14.8 Exercise - Page 978: 43


$\dfrac{L^3}{3\sqrt 3}$

Work Step by Step

Use Lagrange Multipliers Method: $\nabla f(x,y,z)=\lambda \nabla g(x,y,z)$ Volume of a box is given by $f(x,y,z)=V=xyz$ This yields $\nabla f=\lt yz,xz,xy \gt$ and $\lambda \nabla g=\lambda \lt 2x,2y,2z \gt$ Using the constraint condition we get, $yz=\lambda 2x, xz=\lambda 2y,xy=\lambda 2z$ After solving, we get $x=y=z$ Since, $g(x,y,z)=x^2+y^2+z^2=L^2$ yields $x^2+x^2+x^2=L^2$ $x=y=z=\dfrac{L}{\sqrt 3}$ Hence, Volume of a box is given by $f(x,y,z)=V=xyz=(\dfrac{L}{\sqrt 3})^3=\dfrac{L^3}{3\sqrt 3}$
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