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: 45

Answer

$\dfrac{8r^3}{3\sqrt 3}$

Work Step by Step

Volume of a rectangular box is given by $V=abc$ which has the general equation: $x^2+y^2+z^2=4r^2$ This implies $z=\sqrt{4r^2-x^2-y^2}$ This gives $V_x=0, V_y=0$ and $a=b=c=\dfrac{2r}{\sqrt 3}$ Hence, the maximum volume of a rectangular box will be $V=abc=(\dfrac{2r}{\sqrt 3})^3=\dfrac{8r^3}{3\sqrt 3}$
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