Positive numbers $x$ and $y$ satisfy $x + y = 40$. For which value of $y$ is $xy^3$ largest?

Explanation

1. 01

   With $x = 40 - y$, $P = (40 - y)y^3$, so $P' = 3y^2(40 - y) - y^3 = y^2(120 - 4y)$.

2. 02

   $P' = 0$ for positive $y$ at $y = 30$; $P'$ goes from positive to negative there, so it is a maximum, with $x = 10$.