Constrained (and non) extrema of $f(x,y)=2x^2-2xy^3+3y^2$

Question

I need to find the critical points on the boundary,inside $D$,outside $D$, and find the image of this function (constrained on $D$).

$f(x,y)=2x^2-2xy^3+3y^2$

$D=\{2x^2+3y^2\le 9\}$

Critical points non-constrained:

$f_x=4x-2y^3=0$

$f_y=-6xy^2+6y=0$ --> $6y(-xy+1)=0$

$f_y$ is equal to $0$ if $y=0$ or $x=\frac{1}{y}$, plugging it in $f_x$ brings to these critical points : $(0,0),(\pm \frac{1}{2^{\frac{1}{4}}},\pm 2^{\frac{1}{4}})$

using the second derivate test: $(0,0),(\pm \frac{1}{2^{\frac{1}{4}}},\pm 2^{\frac{1}{4}})$ seem to be saddle points (?).

Constrained Critical points (extremas):

$$ \left\{ \begin{array}{c} 4x-2y^3=4x\lambda\\ -6xy^2+6y=6y\lambda\\ 2x^2+3y^2-9=0 \end{array} \right. $$

$-6xy^2+6y=6y\lambda$ --> $6y(-x+1-\lambda)=0$ from this one I can see that it nullifies when $y=0$ or $x=1-\lambda$.

If $y=0$ the first equation gets to $4x(1-\lambda)=0$ which means that it nullifies when $x=0$ or $\lambda = 1$. Looking at the third equation tells us that $(x,y)=(0,0)$ can't be used;

Plugging $y=0$ in the third equation : possible critical points $(\pm \frac{3}{\sqrt(2)},0)$

What I can say about $(\pm \frac{3}{\sqrt(2)},0)$ ?

I think I missed some points.

Answer 1

Critical points non-constrained: $$(0,0),\quad \left(\pm2^{-1/4}, \pm2^{1/4}\right)$$ Global minimum: $$f_{min}=f(0,0)=0$$ Saddle points: $$f\left(\pm2^{-1/4}, \pm2^{1/4}\right)=2\sqrt2$$ Constrained Critical points: from system $$\left\{ \begin{array}{c} 4x-2y^3=4x\lambda\\ -6xy^2+6y=6y\lambda\\ 2x^2+3y^2-9=0 \end{array} \right.$$ eliminate $\lambda$. We get $$ \left\{ \begin{array}{c} (2x^2-y^2)y^2=0\\ 2x^2+3y^2-9=0 \end{array} \right. $$ Solutions is $$\left(\pm\frac{3}{\sqrt2},0\right),\; \left(\frac{3}{2\sqrt2},-\frac32\right),\; \left(-\frac{3}{2\sqrt2},\frac32\right),\; \left(\frac{3}{2\sqrt2},\frac32\right),\; \left(-\frac{3}{2\sqrt2},-\frac32\right) $$ Global maxima: $$f_{max}=f\left(\frac{3}{2\sqrt2},-\frac32\right)=f\left(-\frac{3}{2\sqrt2},\frac32\right)=9+\frac{81}{{{2}^{\frac{7}{2}}}}\approx16.159456$$ Local minimum: $$f_{min}=f\left(\frac{3}{2\sqrt2},\frac32\right)=f\left(-\frac{3}{2\sqrt2},-\frac32\right)=9-\frac{81}{{{2}^{\frac{7}{2}}}}\approx1.84054384$$ Saddle points: $$f\left(\pm\frac{3}{\sqrt2},0\right)=9$$ WolframAlpha return in this case "local maxima".

Other method: parametric equations of ellipse is $x=\frac{3 \cos{(\phi)}}{\sqrt{2}},y=\sqrt{3} \sin{(\phi)}$. We get $$f=9-9 \sqrt{6} \cos{(\phi)} {{\sin{(\phi)}}^{3}}$$ with critical points on $[0, 2\pi]$: $$0,\frac{\pi}{3},\frac{2\pi}{3},\pi,\frac{4\pi}{3},\frac{5\pi }{3}$$

Answer 2

By AM-GM $$2x^2+3y^2-2xy^3\leq9+|2xy^3|=9+\frac{1}{\sqrt2}\sqrt{2x^2(y^2)^3}\leq$$ $$\leq9+\frac{1}{\sqrt2}\sqrt{\left(\frac{2x^2+3y^2}{4}\right)^4}\leq9+\frac{1}{\sqrt2}\sqrt{\left(\frac{9}{4}\right)^4}=9+\frac{81}{16\sqrt2}.$$ The equality occurs for $2x^2=y^2$, $-xy^3=|xy^3|$ and $2x^2+3y^2=9,$ which gives $$(x,y)=\left(-\frac{3}{2\sqrt2},\frac{3}{2}\right),$$ which says that we got a maximal value.

The minimal value is $0$, of course.

In your solution it should be $y=0$, which does not give something with $2x^2+3y^2=9,$ or $$1-xy=\lambda.$$