If x and y are positive real numbers such that their sum is 1, the maximum value of $$x^4y + xy^4$$ is

$$=x^4y + xy^4$$

$$=xy(x^3 + y^3)$$

$$=xy(x + y)(x^2 + y^2 - xy)$$

$$=xy(x^2 + y^2 - xy)$$                                  [Since, x + y = 1]

$$=xy[(x + y)^2 - 2xy - xy)]$$

$$=xy[1 - 3xy]$$

Ley xy = a

$$S=a[1 - 3a]$$

$$S=a - 3a^2$$

$$\frac{\text{d}S}{\text{d}a} = 1 - 6a$$

For maximum; 

$$\frac{\text{d}S}{\text{d}a} = 1 - 6a = 0$$

$$a = xy = \frac{1}{6}$$

After putting $$xy = \frac{1}{6}$$ :

$$=xy[1 - 3xy]$$

$$=\frac{1}{6}(1 - 3 * \frac{1}{6})$$

$$=\frac{1}{12}$$