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Let $f(x_1,\ldots,x_n) = x_1x_2\cdot \cdot \cdot x_n$. Let $A : = \{ x \in \mathbb{R}^n : x_1 + x_2 + \cdots + x_n = n , \; \; x_i \geq 0 \; \; \forall i \} $. I want to find the global max of $f$ under the constrain given by $A$. MY approach:

Put $F(x_1,\ldots,x_n) = \sum_{i=1}^n x_i - n = 0 $ is the constrain. We use lagrange multipliers:

$$\nabla f = \lambda \nabla F \implies (x_2 \cdot \cdot \cdot x_n, \ldots, x_1 \cdots x_{n-1}) = (\lambda,\lambda,\ldots,\lambda).$$ Hence we have

$$ x_2\cdot \cdot \cdot x_n = x_1 x_3 \cdot \cdot \cdot x_n = \cdots = x_1\cdot \cdot \cdot x_{n-1}$$

this implies that $$ x_1 = x_2 = \cdots = x_n $$

We also know $\sum x_i = n \implies n X = n \implies X = 1 $ if we let $X = x_i$

So, max occurs at $(1,1,\ldots,1)$ which is $f(1,1,\ldots,1) = 1$ Is this correct? I feel im missing something. thanks for any feedback.

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maybe for a rigorous argument you should observe that you also have a boundary where clearly there cannot be maximum(because the function is 0 in that case). So since there is a maximum(because your set is compact), and is not in the boundary, it has to be an inner one. So you can go on with your argument.

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