Let $x$ and $y$ be nonnegative real numbers such that $x + y = 1.$  Find the maximum value of
\[x^4 y + xy^4.\]
We can write
\begin{align*}
x^4 y + xy^4 &= xy(x^3 + y^3) \\
&= xy (x + y)(x^2 - xy + y^2) \\
&= xy [(x + y)^2 - 3xy] \\
&= xy (1 - 3xy) \\
&= \frac{3xy (1 - 3xy)}{3}.
\end{align*}By AM-GM,
\[3xy (1 - 3xy) \le \left( \frac{3xy + (1 - 3xy)}{2} \right)^2 = \frac{1}{4},\]so
\[x^4 y + xy^4 \le \frac{1}{12}.\]Equality occurs when $x + y = 1$ and $3xy = \frac{1}{2}.$  By Vieta's formulas, $x$ and $y$ are the roots of $t^2 - t + \frac{1}{6} = 0.$  These roots are
\[\frac{3 \pm \sqrt{3}}{6}.\]Hence, the maximum value is $\boxed{\frac{1}{12}}.$