$\underline{\text { Folklore }}$

Find the maximum value of the expression $x^{2} y-y^{2} x$, if $0 \leq x \leq 1$ and $0 \leq y \leq 1$.

#
$x^{2} y-y^{2} x=x y(x-y)>0$ for $x>y>0$, which means that the maximum value of this expression (if it is achieved) is positive. Therefore, it is sufficient to consider the case $0<y<x \leq 1$.

Under these constraints, according to the Cauchy inequality, $4 x y(x-y) \leq x(y+(x-y))^{2}=x^{3} \leq 1$. Therefore, the maximum value of our expression is $1 / 4$ and is achieved when $x=1, y=x-y=1 / 2$.

## Answer