Let $A= \lbrace (x,y,s,t) \in \mathbb{R}^4 \: | \: x,y,s,t >0 \rbrace$. Find the extreme values for $f:A \to \mathbb{R}$ defined as $f(x,y,s,t)=xyst$ over the constraint $x+y+s+t=4c$ where $c>0$ is fixed.
Again, I tried to attack this problem with the Lagrangian multipliers technology. So I got the following auxiliar function
$$L(x,y,s,t, \lambda )=xyst+ \lambda(x+y+s+t-4c)=xyst+ \lambda x+\lambda y +\lambda s + \lambda t -4 \lambda c.$$
This way, setting
$$ \begin{split} \frac{\partial L}{ \partial x} &= yst + \lambda =0 \\ \frac{\partial L}{ \partial y} &= xst + \lambda =0 \\ \frac{\partial L}{ \partial s} &= xyt + \lambda =0 \\ \frac{\partial L}{ \partial t} &= xys + t =0 \\ \frac{\partial L}{ \partial \lambda} &= x+y+s+t-4c \end{split} $$
So how can I proceed solving the system from here in order to find the values who are the extreme values. For being honest I dont know what role the hypothesis about $c$ and $A$ plays here :(
I already can solve other Lagrangian problems but some like this are giving me problems. Thanks for your help to solve this problem.
I think you made a mistake in calculations. Let $f_x$ be a shorthand notation for $\partial f/\partial x$, then the correct partials are below: $$ \begin{split} 0 &= L_x &= yst + \lambda \\ 0 &= L_y &= xst + \lambda \\ 0 &= L_s &= xyt + \lambda \\ 0 &= L_t &= xys + \lambda \\ 0 &= L_\lambda &= x+y+s+t-4c \end{split} $$ The implication of the first two equations is that $$ yst + \lambda = xst + \lambda \iff (y-x)st = 0 $$ So either $s=0,t=0$ or $y=x$. The first two imply $f=0$ so we are left with $y=x$. Similarly combining all first four equations, we get $x=y=s=t$, and plugging that into the constraint implies $$ x=y=s=t=c \implies f(x,y,s,t) = c^4, $$ which is the attained maximum value of $f$ over the region.
Note that the minimum value of $f=0$ occurs when any of the four independent variables are $0$, and is unattained over the region, since we are only considering strictly positive inputs into $f$.