Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 14 - Section 14.7 - Maximum and Minimum Values - 14.7 Exercise - Page 969: 51

Answer

cube with edge length: $\dfrac{c}{12}$

Work Step by Step

Use Lagrange Multipliers Method: $\nabla f=\lambda \nabla g$ Volume of a box is given by $V(f)=xyz$ $g=4x+4y+4z=c$ This yields $\lt yz,xz,xy \gt =\lambda \lt 4,4,4 \gt$ and $y=x,z=x$ Using the constraint condition we get, $4x+4y+4z=c \implies x=\dfrac{c}{12}$ Hence, the box is a cube with edge length: $\dfrac{c}{12}$
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