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

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.

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 $$[0, 2\pi]$$: $$0,\frac{\pi}{3},\frac{2\pi}{3},\pi,\frac{4\pi}{3},\frac{5\pi }{3}$$

• what did you do to know that $(0,0)$ is a global minimum? – NPLS Feb 11 at 16:50

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 $$1-xy=\lambda.$$

• Can you tell me what I did wrong with Lagrange multipier? – NPLS Feb 11 at 10:59
• @NPLS I added something. I hope it will help. – Michael Rozenberg Feb 11 at 11:38