Maximize $xy^2$ on the ellipse $x^2+4y^2=4$ I was using Lagrange multiplier, any steps gone wrong?
$$f(x,y)=xy^2$$
$$c(x,y)=x^2+4y^2$$
Partial Derivatives
$$\frac {\partial f}{\partial x} = y^2 $$
$$\frac {\partial f}{\partial y} = 2xy $$
$$\frac {\partial c}{\partial x} = 2x $$
$$\frac {\partial c}{\partial y} = 8y $$
Find Lambda
$$y^2=2x\lambda$$
$$\lambda=\frac {y^2}{2x} $$
$$ 2xy=8y\lambda $$
$$\lambda= \frac{x}{4} $$
Let
$$ \frac {y^2}{2x}=\frac {x}{4} $$ yields $$y=\frac {x}{\sqrt 2}$$ and $$y=-\frac {x}{\sqrt 2}$$
Bringing $$y=\frac {x}{\sqrt 2}$$ into C
$$ x^2 + 2x^2 = 4 $$
$$ 3x^2=4 $$ yields $$ x=\frac {2}{\sqrt 3} and -\frac {2}{\sqrt 3} $$
Since $$ y=\frac {x}{\pm\sqrt 2} $$
plugging $x=\frac {2}{\pm\sqrt 3}$ into y I got 4 points:
$(\frac {2}{\sqrt 3},\frac {2}{\sqrt 6}),(-\frac {2}{\sqrt 3},\frac {2}{\sqrt 6}),(-\frac {2}{\sqrt 3},-\frac {2}{\sqrt 6}),(\frac {2}{\sqrt 3},-\frac {2}{\sqrt 6})$
is every step ok up to this point?

You can suppose $x,y>0$ otherwise $xy^2\leq 0$. Then you can use that $\frac{x_1+x_2+x_3}{3}\geq \sqrt[3]{x_1x_2x_3}$ and the equality holds iff $x_1=x_2=x_3$. Note that $(x^2+2y^2+2y^2)^3\geq 27x^2y^4$. So $64\geq 27(xy^2)^2$ the eqaulity holds iff $x=\sqrt{2}y$ or $x=-\sqrt{2}y$. Therefore $xy^2\leq \frac{8}{3\sqrt{3}}$.
