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: 47

Answer

$\dfrac{4}{3} $

Work Step by Step

Volume of a rectangular box is given by $V=xyz$ From the given question, we have $x=6-2y-3z$ Thus, $V=6yz-2y^2z-3yz^2$ This gives $V_y=6z-4yz-3z^2, V_z=6y-2y^2-6yz$ After solving, we get $y=1,z=\dfrac{2}{3}$ and $x=6-2y-3z=2$ Hence, the volume of a rectangular box will be $V=xyz=(2)(1)(\dfrac{2}{3})=\dfrac{4}{3} $
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