A non-linear maximisation

Question

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 $2$?

Answer 1

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)

Answer 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 $(x-2)^2(x+1) \geq 0$.

Answer 3

$x^2y=x^2(3-x)=3x^2-x^3$

So, $\frac{x^2y}{dx}=3(2x)-3(x^2)=3x(2-x)$

For the extreme values of $x^2y,\frac{x^2y}{dx}=0\implies x=0$ or $x=2$

Now, $\frac{d^2(x^2y)}{dx^2}=6-3(2x)=6(1-x)$

At $x=0,\frac{d^2(x^2y)}{dx^2}=6>0$ so $x^2y$ will have minimum value at $x=0$

At $x=2,\frac{d^2(x^2y)}{dx^2}=-6<0\implies x=2$ will make $x^2y$ the maximum

Answer 4

Besides' to @lab's answer, the following fact may help you:

If $x$ and $y$ be positive numbers such that $ax+by=k$, $k$ is constant, then $x^{\alpha}y^{\beta}$ has maximum value if $\frac{ax}{\alpha}=\frac{by}{\beta}=\frac{k}{\alpha+\beta}$.

Now find the value.