We simplify the given expression: $\begin{aligned} x^4y + xy^4 &= xy(x^3+y^3) \\ &= xy(x+y)(x^2-xy+y^2) \\ &= xy(x^2-xy+y^2) \\ &= xy[(x+y)^2 - 3xy] \\ &= xy(1-3xy). \end{aligned}$ Now let $t = xy,$ so the expression becomes simply $t(1-3t) = -3t^2 + t.$

The maximum value of this expression occurs when $t = -\dfrac{1}{2(-3)} = \dfrac16.$ Luckily, this value of $t$ is attainable: solving the system $x + y = 1$ and $xy = \dfrac16$ gives the solution $(x, y) = \left(\dfrac{3-\sqrt3}{6}, \dfrac{3+\sqrt3}{6}\right)$ in some order.

Thus, the maximum value of the expression is $-3 \cdot \left(\dfrac16\right)^2 + \dfrac16 = \boxed{\dfrac1{12}}.$ (We would enter the approximation $0.083.$ )

$x^{4}y+xy^{4}=xy(x^{3}+y^{3})=xy(x+y)(x^{2}-xy+y^{2})$ $=xy((x+y)^{2}-3xy)=xy(1-3xy)=\frac{3xy(1-3xy)}{3}$

By using Cauchy's inequality: $\frac{3xy(1-3xy)}{3}≤\frac{(\frac{3xy+1-3xy}{2})^{2}}{3}=\frac{1}{12}$

Here's the simple solution:

We got the degree of given equation as 5, thus I took 1's fifth part i.e. 0.2 as x and so as per condition y =0.8

Put these values to get the answer as 0.0832 precisely.

P.S. :

For practice you can try various other functions of degree 2, 3,4 and can get accurate results with same method. I know this is shortcut, but it is still effective.

Set $(x,y)=(\cos^2 t, \sin^2t$ where $t\in (0,\pi)-\{\pi/2\}$ , then : $\begin{aligned} x^4y+xy^4 &= \cos^8 t (1-\cos^2 t) +(1-\cos^2 t)^4\cos^2 t \\ &= \cos^2 t-4\cos^{4}t+6\cos^6 t -3 \cos^8 t \\ &= \frac{7-4\cos (4t) -3 \cos(8t)}{128} \\ &\overset{y=\cos 4t}{=} \frac{5-2y-3y^2}{64}\end{aligned}$ Now the problem became really simple, we just need to find $\min\left\{\frac{5-2y-3y^2}{64}\ |\ -1\leq y \leq 1 \right\}$ , which is easy : $\frac{1}{64}(5-2y-3y^2) = \frac{1}{64} \left(\frac{16}{3} - \left(y\sqrt{3} +\frac{1}{\sqrt{3}} \right)^2 \right) \leq \frac{1}{64}\cdot \frac{16}{3} = \frac{1}{12}$ Equality is attainable at $y=\frac{-1}{3} \in (-1,1)$ then the maximum is $\frac{1}{12} \approx 0.08333..$

While what I did was similar to many others, I am wondering why x=y gave us neither the maximum nor the minimum value?