5. Find the maximum possible value of the expression $x y(x-y)$, given that $x, y \in$ $[0 ; 1]$
Answer: 0.25.

Solution: Consider $x y(x-y)=-x y^{2}+x^{2} y$ as a quadratic function of $y$. Its maximum is achieved at the vertex of the corresponding parabola $y_{0}=\frac{x}{2}$. Clearly, it is located on the given segment. Substituting into the expression, we get $-x y_{0}^{2}+x^{2} y_{0}=\frac{x^{3}}{4}$. Clearly, the maximum value is $\frac{1}{4}$.