## Calculus: Early Transcendentals 8th Edition

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