Let $f:\mathbb{R}^n\rightarrow \mathbb{R}\;$ be defined as follows:
$\forall x\in \mathbb{R}^n \; f(x)= x_1x_2...x_n$
I need to find a point $a = (a_1,...a_n) \in \mathbb{R}^n\;$ such that $a$ is a critic point of $f$ under the following restrictions:
$\forall 1\leq i \leq n \; a_i \neq 0$
$x_1 + ... + x_n = n$
I started by calculating the gradient of $f$:
$\nabla f=\left(\begin{array}{c} x_{2}x_{3}...x_{n}\\ x_{1}x_{3}...x_{n}\\ \vdots\\ x_{1}x_{2}...x_{n-1} \end{array}\right)$
then I defined $g(x) = x_1 + x_2 + ... + x_n -n \;$ as my restriction function, we have that:
$\nabla g=\left(\begin{array}{c} 1\\ 1\\ \vdots\\ 1 \end{array}\right)$
I was having trouble solving the system $\nabla f = \lambda \nabla g\;$ under the above restrictions though.
Would love to get some help.
From $\nabla f=\lambda\nabla g$ you obtain a system of $n$ equations with high symmetry. The first equation is $$x_2x_3\ldots x_n=\lambda\ .$$ Multiply this equation with $x_1$, and obtain $$p:=\prod_ix_i=\lambda x_1\ .$$ In the same way you obtain $n$ equations $$p=\lambda x_j\qquad(1\leq j\leq n)\ .$$ This allows to conclude that a conditionally critical point with all $x_i\ne0$ has $x_1=x_2=\ldots=x_n$, and therefore $x_i=1$ $\>(1\leq i\leq n)$.