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?

$$f = xy^2$$
with
$$4=x^2+4y^2$$
becomes
$$f = xy^2 = x\left(1-\frac{1}{4}x^2\right) = x - 0.25x^3$$
maximum via derivative
$$\frac{d}{dx}f = 0 = 1 - 0.75x^2$$
$$x_{1,2}=\pm\sqrt{\frac{4}{3}}$$
$$\frac{d^2}{dx^2}f(x_1) < 0 $$
$$\frac{d^2}{dx^2}f(x_2) > 0 $$
and with $4=x^2+4y^2$ again:
$$y_{1,2} = \pm\sqrt{\frac{2}{3}}$$
I think that's somewhat plausible, as the $y^2$ creates a symmetry around the x axis. The two points are $(x_1,y_1)$ and $(x_1,y_2)$
