# extrema and range of a multivariable function (constrained)

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

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 $$(0,0,0)$$,$$(0,y,0)$$)?

• Under the constraint $Im(f)$ is $(0,\frac 1 3]$. – Kavi Rama Murthy Feb 12 at 11:43
• It seem $\frac{1}{3}$ is in outer space. – Takahiro Waki Feb 12 at 12:20
• @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 $x^2-y^2+z^2=3$ – NPLS Feb 12 at 12:43