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: 50

Answer

Dimensions: $x=\sqrt[3] {\dfrac{2}{5}}V$,$y=\sqrt[3] {\dfrac{2}{5}}V$ and $z=\dfrac{5}{2}\sqrt[3] {\dfrac{2}{5}}V$ or, Dimensions: $\sqrt[3] {\dfrac{2}{5}}V \times \sqrt[3] {\dfrac{2}{5}}V \times \dfrac{5}{2}\sqrt[3] {\dfrac{2}{5}}V$

Work Step by Step

Use Lagrange Multipliers Method: $\nabla f=\lambda \nabla g$ Volume of a box is given by $V=xyz$ This yields $\nabla f=\lt 10x+4z,4x \gt$ and $\lambda \nabla g=\lambda \lt 2xz, x^2 \gt$ Using the constraint condition we get, $10x+4z=\lambda 2xz, 4x=\lambda x^2,V=x^2z$ After solving, we get $z=\dfrac{5x}{2}$ and $V=x^2z$ yields $x=\sqrt[3] {\dfrac{2}{5}}V$ Hence, Dimensions: $x=\sqrt[3] {\dfrac{2}{5}}V$,$y=\sqrt[3] {\dfrac{2}{5}}V$ and $z=\dfrac{5}{2}\sqrt[3] {\dfrac{2}{5}}V$ or, Dimensions: $\sqrt[3] {\dfrac{2}{5}}V \times \sqrt[3] {\dfrac{2}{5}}V \times \dfrac{5}{2}\sqrt[3] {\dfrac{2}{5}}V$
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