Maximize $xy^2$ on the ellipse $x^2+4y^2=4$

Question

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?

Answer 1

$$f = xy^2$$

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 $(x_1,y_2)$

Answer 2

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 $xy^2\leq \frac{8}{3\sqrt{3}}$.