Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 11 - Cumulative Review Exercises - Page 1182: 39



Work Step by Step

Let $ y $ be the height of the box. Surface Area, $ A=x^2+4xy $ $ V=x^2y=4 \implies y=\dfrac{V}{x^2}=\dfrac{4}{x^2}$ Surface Area, $ A=x^2+4xy=x^2+4x \times \dfrac{4}{x^2}$ or, $ A=x^2+\dfrac{16}{x}=\dfrac{x^3+16}{x}$
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