Maximising $f(x,y)=xy^2$ subject to $x^2+y^2 = 1$ using the Lagrange Multiplier As the title suggests, I'm looking for the maximum of $f(x,y)=xy^2$ on 
$$ E:= \{(x,y)\in \mathbb{R}^2| x^2+y^2 \leq 1\} $$
As $E$ is a compact I already know $f$ will have a maximum and a minimum there. Because the exercise suggested I use Lagrange Multipliers I'm guessing they are going to lie on the edge where $x^2+y^2 = 1$. Also, I believe, if I find the maximum and minimum there I need not search for additional ones as they would be unique.
Therefore, I define $g(x):= x^2+y^2 - 1$ such that the roots of the functions are the set $Ng:=\{(x,y)\in \mathbb{R}^2| \quad||(x,y)||_2^2=1\}$ and this set is also compact. Furthermore, both $f$ and $g$ are continuously differentiable.
Now because $$\nabla g(x,y)=(2x,2y)\neq 0$$ for values in $Ng$, there exists $\lambda \in \mathbb{R}$ such that 
$$ \nabla f(x,y)=\lambda \nabla g(x,y)  $$
But can somebody help me make sense of the equivalent system
$$ y^2 = \lambda 2x $$
$$ 2xy = \lambda 2y $$ 
How do I find the maximum/ minimum points that I'm looking for from this system of equations?
 A: If $x=0$ or $y=0$ then it is easy to find the point.
Otherwise, dividing the two equations you get
$$\frac{y^2}{2xy}=\frac{\lambda 2x}{\lambda 2y} \Rightarrow \frac{y}{2x}=\frac{x}{y} \Rightarrow y^2=2x^2$$
You also know that $x^2+y^2=1$.
P.S. If $x_0 <0$ then $f$ cannot have a maximum at $(x_0, y_0)$. And if $x_0 >0$ then it is easy to check that 
$$f(x_0,y_0) < f(x_0+\epsilon, y_0+\epsilon)$$
This implies that the maximum is indeed achieved on the circle $x^2+y^2=1$.
A: By AM-GM
$$xy^2\leq\sqrt{x^2y^4}=\sqrt{\frac{1}{2}\cdot2x^2\cdot y^2\cdot y^2}\leq\sqrt{\frac{1}{2}\left(\frac{2x^2+y^2+y^2}{3}\right)^3}=\frac{2}{3\sqrt3}.$$
The equality occurs for $x=\frac{1}{\sqrt3}$ and $y=\sqrt{\frac{2}{3}},$ which says that the answer is $\frac{2}{3\sqrt3}.$
Done!
A: Here's another approach, wich leads to simpler computations. Since we are working in the region $x^2+y^2=1$, you can replace $f$ by the function $(x,y)\mapsto x(1-x^2)=x-x^3$. So, applying the method of Lagrange multipliers, you get the system$$\left\{\begin{array}{l}1-3x^2=2\lambda x\\0=2\lambda y\\x^2+y^2=1.\end{array}\right.$$Therefore, $y=0$ or $\lambda=0$. If $y=0$, you have the system$$\left\{\begin{array}{l}1-3x^2=2\lambda x\\x^2=1.\end{array}\right.$$In this case, $x=\pm1$. Otherwise, you have the system$$\left\{\begin{array}{l}1-3x^2=0\\x^2+y^2=1.\end{array}\right.$$In this case, the solutions are $\pm\left(\sqrt{\frac13},\sqrt{\frac23}\right)$ and $\pm\left(\sqrt{\frac13},-\sqrt{\frac23}\right)$.
A: Did you forget to differentiate with respect to $\lambda$? It just gives you the constraint. With this constraint there are three equations in three unknowns:
$$\begin{align}y^2=\lambda 2x\end{align}\tag1$$
$$\begin{align}2xy=\lambda 2y\end{align}\tag2$$
$$\begin{align}x^2+y^2=1\end{align}\tag3$$
Let's make a univariate polynomial equation from these equations. Multiply $(1)$ by $x\neq 0$, $(2)$ by $y\neq 0$, add and use $(3)$
$$\begin{align}2\lambda=xy^2+2xy^2=3xy^2\end{align}$$
Use $xy=\lambda y$ and then $(1)$ again
$$\Leftrightarrow$$
$$\begin{align}2\lambda=3xy^2=3\lambda y^2=6\lambda^2x\end{align}$$
Divide both $2$, multiply both sides by $y\neq 0$ and use $xy=\lambda y$
$$\Leftrightarrow$$
$$\begin{align}\lambda y=3\lambda^2xy=3\lambda^3y\end{align}$$
$$\Leftrightarrow$$
$$\begin{align}(3\lambda^3-\lambda)y=0\end{align}$$
Since $y\neq 0$ it must be that $3\lambda^3-\lambda=0$. Then either $\lambda=0$ or $3\lambda=-\sqrt 3$ or $3\lambda=+\sqrt 3$. In the first case $x=\pm 1$. In the second $x=-\sqrt{\dfrac{1}{3}}$ and the third $x=+\sqrt{\dfrac{1}{3}}$. 
Maybe my algebra is inelegant here. Suggestion are welcome.
