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

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

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 ?

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  • $\begingroup$ 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? $\endgroup$ – maxmilgram Feb 11 at 15:21
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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 $r$.

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