A non-linear maximisation We know that $x+y=3$ where x and y are positive real numbers. How can one find the maximum value of $x^2y$? Is it $4,3\sqrt{2}, 9/4$ or $2$?

By AM-GM 
$$\sqrt[3]{2x^2y} \leq \frac{x+x+2y}{3}=2 $$
with equality if and only if $x=x=2y$.
Second solution
This one is more complicated, and artificial (since I needed to know the max)$.
$$x^2y=3x^2-x^3=-4+3x^2-x^3+4=4- (x-2)^2(x+1)\leq 4$$
since $(x-2)^2(x+1) \geq 0$.
