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: 10

Answer

$P$ has a maximum value of $P(2)=20$.

Work Step by Step

We need to maximize $P$. $P(I)=\frac{100I}{I^{2}+I+4}$, so $P'(I)=\frac{-100(I+2)(I-2)}{(I^{2}+I+4)^{2}}$. The derivative is positive for $[0,2]$ and negative for $[2,\infty)$, so $P$ is maximized at $P(2)=20$.
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