BMO maximum

Using AM-GM: $x^2y-xy^2=xy(x-y)≤\left(\frac{xy+x-y}{2}\right)^2=\frac{\left((x-1)(y+1)+1\right)^2}{4}$

We then see that $(x-1)(y+1)≤0$ for $0≤x,y≤1.$

So the max value occurs at $\frac{1}{4}$ when $x=1$ and $y=\frac{1}{2}$

(x^2)y-x(y^2) = xy(x-y). This can be maxed by keeping at it uppermost level i.e. x= 1 This reduces the function to y(1-y) whose maximum value comes at y = 1/2 and the value of the function then is 1/4 = m/n Therefore, m+n = 5

Same solutions.

Let $A=x^2y-y^2x=xy(x-y)$ .

If $x<y$ , then $A<0$ .

If $x>y$ , applying AM-GM inequality we have:

$A=xy(x-y)\le \dfrac{x(y+x-y)^2}{4}=\dfrac{x^3}{4}\le\dfrac{1}{4}$

The equality holds if and only if $x=1; y=\dfrac{1}{2}$ .

So, the maximum value of $A$ is $\dfrac{1}{4}$ and $m+n=\boxed{5}$ .