Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 3 - Applications of Differentiation - 3.7 Optimization Problems - 3.7 Exercises - Page 266: 39

Answer

Diameter $16$ $in.$

Work Step by Step

The perimeter of a slice is $2r+r\theta=32$ so we have: $\theta=\frac{32-2r}{r}$. The area $A$ of the slice is $A$ = $\frac{1}{2}{r^{2}}{θ}$ = $\frac{1}{2}{r^{2}}\left(\frac{32-2r}{r}\right)$ = $16r-r^{2}$ for $0$ $\leq$ $r$ $\leq$ $16$ $A'(r)$ = $16-2r$ $A'(r)$ = $0$ $16-2r$ = $0$ $r$ = $8$ Since $A(0)$ = $0$, $A(16)$ = $0$ and $A(8)$ = $64$, the largest piece comes from a pizza with radius $8$ $in.$ and diameter $16$ $in.$ $θ$ = $2$ radians $\approx$ $114.6°$ which is about $32$% of the whole pizza.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.