jmc

We can write $$\begin{array}{rl}{x}^{4}y+x{y}^4& =xy(x^3+y^3)\\ & =xy(x+y)(x^2-xy+y^2)\\ & =xy[(x+y{)}^2-3xy]\\ & =xy(1-3xy)\\ & =\frac{3xy(1-3xy)}{3}.\end{array}$$By AM-GM, $$3xy(1-3xy)\le {\left(\frac{3xy+(1-3xy)}{2}\right)}^2=\frac{1}{4},$$so $$x^{4}y+x{y}^4\le \frac{1}{12}.$$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 $\boxed{\frac{1}{12}}.$