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: 40


Dimensions of a box: $x=y=z=\sqrt{\dfrac{32}{3}}$

Work Step by Step

Given: Surface area =64 $cm^2$ Need to apply Lagrange Multipliers Method to determine the maximum volume of a rectangular box. we have $\nabla f=\lambda \nabla g$ The volume of a box is $V=xyz$ Surface area, $S=2xy+2yz+2zx$ From the given question, $S=64 cm^2$ Now, $\nabla V=\lambda \nabla S$ $\lt yz,xz,xy \gt =\lambda \lt 2(y+z), 2(x+z),2(x+y) \gt$ and $xyz=2 \lambda (xz+yz)$ Simplify to get the values of x,y and z. We have $x=y=z=\sqrt{\dfrac{32}{3}}$ Thus, The Dimensions of a box: $x=y=z=\sqrt{\dfrac{32}{3}}$
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