10.3. Find the minimum and maximum values of the expression $3 x^{2} y-2 x y^{2}$, where $x, y$ take arbitrary values from the interval $[0,1]$
Answer: $-\frac{1}{3}$ and $\frac{9}{8}$

Solution 1. Let $f(x, y)=3 x^{2} y-2 x y^{2}$, then $f(x, y)=x y(3 x-2 y)$. The obtained expression equals 0 when $x=0$, or $y=0$, or $y=\frac{3}{2} x$. Therefore, in the region $0 \leq x, y \leq 1$, the function $f(x, y)$ is positive below (or to the right of) the line $y=\frac{3}{2} x$ and reaches its maximum here, and it is negative above (or to the left of) this line and reaches its minimum here. In the first case, due to the inequality $x \leq 1$ and the positivity of the expression in parentheses, we get $f(x, y)=x y(3 x-2 y) \leq y(3-2 y)=3 y-2 y^{2}$. The maximum of this quadratic trinomial is achieved at $y=\frac{3}{4}, x=1$ and is equal to $\frac{9}{8}$, which gives the maximum value of the expression from the condition.

In the second case, due to the inequality $y \leq 1$ and the negativity of the expression in parentheses, we get $f(x, y)=x y(3 x-2 y) \geq x(3 x-2)=3 x^{2}-2 x$. The minimum of this quadratic trinomial is achieved at $x=\frac{1}{3}, y=1$ and is equal to $-\frac{1}{3}$, which gives the minimum value of the expression from the condition.

Solution 2. Consider the variable $y$ as a parameter lying in the interval $[0,1]$. For each fixed $y \in[0,1]$, the quadratic trinomial $f(x)=3 y x^{2}-2 y^{2} x$ takes the minimum value, equal to $-\frac{y^{3}}{3}$, at $x=-\frac{-2 y^{2}}{2 \cdot 3 y}=\frac{y}{3} \in[0,1]$, therefore, the minimum value of the entire expression is achieved at $y=1, x=\frac{1}{3}$ and is equal to $-\frac{1}{3}$. The maximum of the quadratic trinomial $f(x)=3 y x^{2}-2 y^{2} x$ for each fixed $y \in[0,1]$ on the interval $[0,1]$ is achieved at its end at $x=1$ (the value at the other end $x=0$ is 0), and is equal to $3 y-2 y^{2}$. Therefore, the maximum of $f(x)$ does not exceed the maximum of the trinomial $3 y-2 y^{2}$, which is achieved at $y=\frac{3}{4}$ in the interval $[0,1]$ and $x=1$, and is equal to $\frac{9}{8}$.

Grading criteria. One of the estimates $-\frac{1}{3}$ and $\frac{9}{8}$ is proven: 3 points.