Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 14 - Partial Derivatives - 14.7 Exercises - Page 979: 48

Answer

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

Work Step by Step

Volume of a box is given by $V=xyz$ Surface area, $S=2xy+2yz+2zx=64 cm^2$ Use Lagrange Multipliers Method: $\nabla f=\lambda \nabla g$ This yields $\lt yz,xz,xy \gt =\lambda \lt 2(y+z), 2(x+z),2(x+y) \gt$ and $xyz=2 \lambda (xz+yz)$ After solving we get, $x=y=z$ Since, $S=2xy+2yz+2zx=64 cm^2$ Hence, Dimensions of a box are: $x=y=z=\sqrt{\dfrac{32}{3}}$
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