For real numbers $x, y$, it is given that $x+y=1$. Determine the maximum value of the expression $A(x, y)=x^{4} y+x y^{4}+x^{3} y+x y^{3}+x^{2} y+x y^{2}$.
I. solution. Let's perform the following transformations:

$$
\begin{aligned}
A(x ; y) & =x y(x^{3}+y^{3}+x^{2}+y^{2}+\underbrace{x+y}_{1})= \\
& =x y[(\underbrace{x+y}_{1})\left(x^{2}-x y+y^{2}\right)+x^{2}+y^{2}+1]= \\
& =x y(\underbrace{x^{2}+2 x y+y^{2}}_{1}+x^{2}-3 x y+y^{2}+1)=x y\left(x^{2}-3 x y+y^{2}+2\right)= \\
& =x y(\underbrace{x^{2}+2 x y+y^{2}}_{1}-5 x y+2)=x y(3-5 x y) .
\end{aligned}
$$

By introducing the notation $a=x y$, we get $A(x ; y)=a(3-5 a)$.

This is a quadratic expression in $a$ that attains its maximum value at $a=\frac{3}{10}$, and it is strictly increasing in the interval $\left.]-\infty, \frac{3}{10}\right]$. Given the condition $x+y=1$, the maximum possible value of $a$ is $\frac{1}{4}$, which is attained when $x=y=\frac{1}{2}$. Since $\frac{1}{4}<\frac{3}{10}$, we can conclude that the maximum value of the expression $A(x ; y)$ is

$$
A\left(\frac{1}{2} ; \frac{1}{2}\right)=\frac{1}{4}\left(3-\frac{5}{4}\right)=\frac{7}{16}
$$

II. solution. Substitute $(1-x)$ for $y$ in $A$:

$$
\begin{aligned}
A(x ; 1-x)= & f(x)=x^{4}(1-x)+x(1-x)^{4}+x^{3}(1-x)+x(1-x)^{3}+ \\
& +x^{2}(1-x)+x(1-x)^{2}, \quad \text { i.e. } \\
f(x)= & -5 x^{4}+10 x^{3}-8 x^{2}+3 x
\end{aligned}
$$

The expression can have an extremum where its first derivative is 0:

$$
f^{\prime}(x)=-20 x^{3}+30 x^{2}-16 x+3=0
$$

which can also be written as:

$$
\left(x-\frac{1}{2}\right)\left(-20 x^{2}+20 x-6\right)=0
$$

The equation has only $x=\frac{1}{2}$ as a root (the second factor has no real roots). The maximum will occur if the second derivative is negative at this value:

$$
f^{\prime \prime}(x)=-60 x^{2}+60 x-16
$$

Substituting $x=\frac{1}{2}$:

$$
(-60) \cdot \frac{1}{4}+60 \cdot \frac{1}{2}-16=-1
$$

Thus, the expression has a maximum at $x=\frac{1}{2}$. Given $x+y=1$, we have $y=\frac{1}{2}$.

The value of the original expression at this point is: $A\left(\frac{1}{2} ; \frac{1}{2}\right)=\frac{7}{16}$.