Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 1: Functions - Section 1.1 - Functions and Their Graphs - Exercises 1.1 - Page 11: 11


a=d/$\sqrt 3$ surface area=$2d^{2}$ volume=$d^{3}$/3$\sqrt 3$

Work Step by Step

Applying Pythagorean theorem, edge diagonal,D= $\sqrt 2$ a. Applying Pythagorean theorem again will yield body diagonal,d=$ \sqrt 3$a [$d^{2}$=$a^{2}$+($\sqrt 2a$)^2] Now we have a as function of d. surface area=6$a^{2}$ volume=$a^{3}$ Substitute a in terms of d in the above two equations.
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