# Extrema min-max over constrained $f(x,y,z)$

$$f(x,y,z)=xe^{yz}$$

$$D=\{x^2+y^2+z^2\le 25,3x^2+y^2+z^2=27\}$$ $$\left\{ \begin{array}{ll} f_x=e^{yz}=0\\ f_y=xze^{yz}=0\\ f_x=xye^{yz}=0 \end{array} \right.$$

from here I can't see any critical point.

Now in order to find the critical point constrained on $$D$$, I'll use the Lagrange multiplier method:

$$\left\{ \begin{array}{ll} e^{yz}=zx\lambda+6x\mu\\ xze^{yz}=2y\lambda+2y\mu\\ xye^{yz}=2z\lambda+2z\mu\\ x^2+y^2+z^2-25=0\\ 3x^2+y^2+z^2-27=0 \end{array} \right.$$

From that system I found :

$$(\pm1,\sqrt{12},\sqrt{12}),(\pm1,-\sqrt{12},-\sqrt{12}),(\pm1,\sqrt{12},-\sqrt{12}),(\pm1,-\sqrt{12},\sqrt{12})$$

pugging them back into the function :

$$f(1,\pm\sqrt{12},\pm\sqrt{12})=e^{12}$$

$$f(-1,\pm\sqrt{12},\mp\sqrt{12})=-e^{12}$$

$$f(1,\pm\sqrt{12},\pm\sqrt{12})=e^{-12}=\frac{1}{e^{12}}$$

Questions:

1)

Can I state that $$f(1,\pm\sqrt{12},\pm\sqrt{12})=e^{12}$$,$$f(1,\pm\sqrt{12},\pm\sqrt{12})=e^{-12}=\frac{1}{e^{12}}$$ are a maximum , $$f(-1,\pm\sqrt{12},\mp\sqrt{12})=-e^{12}$$ a minimum , without using The $$3x3$$ hessiam matrix ?

2)

Judging by the fact that I didn't find any critical points by setting $$f_x=0,f_y=0,f_z=0$$ does that mean there are not critical points (max-min) inside $$D$$ or outside it, so the only critical points are only those constrained that I found before ?

• Did you find any critical points with just the sharp constraint (the second one) that still are in $D$? So with just one lagrange multiplier? – maxmilgram Feb 11 at 15:21

The set $$D$$ is a rotational elliptical surface, bounded by two circles. There are no $$3$$-variables critical points to consider.
Write $$y=r\cos\phi,\quad z=r\sin\phi\ .$$ Then we have to extremise the function $$g(x,r,\phi):=xe^{r^2\sin(2\phi)/2}$$ under the constraints $$x^2+r^2\leq25,\quad 3x^2+r^2=27\ .\tag{1}$$ Given $$x>0$$ and $$r\geq0$$ the function $$\phi\mapsto g(x,r,\phi)$$ is maximal when $$\sin(2\phi)=1$$. We therefore have to maximize $$g_\max(x,r):=xe^{r^2/2}$$ under the constraints $$(1)$$ and $$x>0$$, $$r\geq0$$. It is then clear that the minimum of $$f$$ is obtained as $$g_\min=-g_\max$$ by replacing $$x_\max$$ by $$x_\min=-x_\max$$, keeping the same $$r$$.