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: 55


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

Work Step by Step

Need to apply 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 $\nabla f=\lt yz,xz,xy \gt$ and $\lambda \nabla g=\lambda \lt 2x,2y,2z \gt$ Using the constraint condition we get, $yz=\lambda 2x, xz=\lambda 2y,xy=\lambda 2z$ Simplify to get the value of $x,y$ and $z$ We have $x=y=z$ From the given question, let us consider $g(x,y,z)=x^2+y^2+z^2=L^2$ yields $x^2+x^2+x^2=L^2$ Thus, $x=y=z=\dfrac{L}{\sqrt 3}$ Therefore, the volume of a box is $V=xyz=(\dfrac{L}{\sqrt 3})^3$ $V=(\dfrac{L}{\sqrt 3})(\dfrac{L}{\sqrt 3})(\dfrac{L}{\sqrt 3})$ $V=\dfrac{L^3}{3\sqrt 3}$
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