0
$\begingroup$

$f(x,y,z)=\frac{1}{3+x^2+z^2}$

Constrained on : $D=\{x^2+z^2\le 3+y^2\}$

Function not-constrained:

$$\displaystyle \left\{ \begin{array}{c} f_x=\frac{2x}{(3+x^2+z^2)^2}=0\\ f_y=0\\ f_z=\frac{2z}{(3+x^2+z^2)^2}=0 \end{array} \right. $$

From here I can say that we have critical points at $(0,y,0)$. It seems also that we have a maximum at $(0,0,0)=\frac{1}{3}$

Function constrained:

Using Lagrange multiplier method : $$\displaystyle \left\{ \begin{array}{c} \frac{2x}{(3+x^2+z^2)^2}=2x\lambda\\ 0=-2y\lambda\\ \frac{2z}{(3+x^2+z^2)^2}=2z\lambda\\ x^2-y^2+z^2-3=0 \end{array} \right. $$

From here I Found these points : $(0,0,\pm \sqrt{3}),(\pm\sqrt{3},0,0)$

The range (Image) of this function constrained should be : $Im(f)=[0,\frac{1}{6}]$

Questions :

Did I do right? Is that everything I can say about Its extremes?

Are $(0,y,0)$ points part of the answer (Can I say that the critical points are $(0,0,0)$,$(0,y,0)$)?

$\endgroup$
  • $\begingroup$ Under the constraint $Im(f)$ is $(0,\frac 1 3]$. $\endgroup$ – Kavi Rama Murthy Feb 12 at 11:43
  • $\begingroup$ It seem $\frac{1}{3}$ is in outer space. $\endgroup$ – Takahiro Waki Feb 12 at 12:20
  • $\begingroup$ @KaviRamaMurthy I plugged $(0,0,\pm \sqrt{3})$ into the function, which gives me $\frac{1}{6}$. Why should I stick with $\frac{1}{3}$ ? $(0,0,0)$ does not satify the constrain $x^2-y^2+z^2=3$ $\endgroup$ – NPLS Feb 12 at 12:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.