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 $2$?

Lagrange multipliers: Let $f(x,y)=x^2y$ and $g(x,y,\lambda)=x^2y-\lambda (x+y-3)$. Then:
$$\partial _y g=0\iff x^2-\lambda=0\iff x=\pm \sqrt{\lambda}$$
$$\partial _x g=0\iff 2xy-\lambda=0\iff xy=\frac{\lambda}2\iff y=\pm \frac{\sqrt{\lambda}}2$$
$$\partial _{\lambda} g=0\iff x+y=3$$
with $\lambda>0$. We want $x+y=3$ and so $\sqrt{\lambda}=2\iff \lambda=4$ or $-\sqrt{\lambda}=2$ which is impossible. Thus,
$$(x,y)=(2,1)$$
and $f(2,1)=4$ is a possible maximum value. To verify that it is the maximum value you will have to determine the eigenvalues of the Hessian (which is simple)
