Calculus: Early Transcendentals 8th Edition

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

Chapter 4 - Section 4.7 - Optimization Problems - 4.7 Exercises - Page 338: 39


The diameter should be $~~16~in$

Work Step by Step

Let $\theta$ be the central angle of the pizza slice. Let $r$ be the radius of the pizza. We can use the perimeter to find an expression for $\theta$: $r+r+\theta~r = 32$ $\theta = \frac{32-2r}{r}$ We can find an expression for the area of the pizza slice: $A = (\frac{\theta}{2\pi})~\pi~r^2$ $A = \frac{\theta~r^2}{2}$ $A = \frac{(32-2r)~r^2}{2r}$ $A = (16-r)~r$ $A = 16r-r^2$ We can find $\frac{dA}{dr}$: $\frac{dA}{dr} = 16-2r = 0$ $r = 8$ Since the diameter is twice the radius, the diameter should be $~~16~in$
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