Lagrange multiplier for sum of products with sum constraint

Question

Is it possible to use Lagrange multipliers (or another technique) to easily find a maximum of a function like $$ f: \begin{cases} \mathbb{R}^3_{\ge0}&\to \mathbb{R}_{\ge0}\\ (x_1,x_2,x_3)&\mapsto x_1x_2x_3+ 2x_1x_2+3x_1x_3+x_2 \end{cases} $$ with the constraint that the sum of the arguments $\sum_{i=1}^3x_i=s$ for some $s\in\mathbb{R}_{>0}$. I know that the product $x_1x_2x_3$ is maximized when all variables are the same but here, we have a weighted sum of partial products. (The function is given only as an example, I actually want to find a maximum of a function with more variables but of the same kind, i.e. weighted sum of partial products where each variable has either power 0 or 1 in the exponent)

Answer 1

According to the Lagrange's Multipliers technique the formulation could be

Given

$$ f(x) = x_1 x_2 x_3 +2x_2 x_2+3x_1 x_3+x_2\\ g_1(x) = x_1+x_2+x_3 - s = 0\\ g_2(x,\epsilon) = x_1 - \epsilon_1^2 = 0\\ g_3(x,\epsilon) = x_2 - \epsilon_2^2 = 0\\ g_4(x,\epsilon) = x_3 - \epsilon_3^2 = 0\\ $$

Determine the stationary points for

$$ L(x,\lambda,\epsilon) = f(x)+\lambda _1g_1(x)+\sum_{k=1}^3\lambda_{k+1}g_{k+1}(x_k,\epsilon_k) $$

Here $\lambda_k$ are the so called Lagrange multipliers and $\epsilon_k$ are slack variables to transform the restrictions $x_k \ge 0$ into equivalent equality restrictions.

Now the stationary points are the solutions for

$$ \nabla L = \left\{ \begin{array}{rcl} \lambda_1+\lambda_2+2 x_2+x_2x_3+3 z=0 \\ \lambda_1+\lambda_3+2 x_1+x_1 x_3+1=0 \\ \lambda_1+\lambda_4+3 x_1+x_1 x_2=0 \\ x_1+x_2+x_3-s=0 \\ x_1-\epsilon_1^2=0 \\ x_2-\epsilon_2^2=0 \\ x_3-\epsilon_3^2=0 \\ -2 \epsilon_1 \lambda_2=0 \\ -2 \epsilon_2 \lambda_3=0 \\ -2 \epsilon_3 \lambda_4=0 \\ \end{array} \right. $$

After that, the solutions should be qualified as local minimum, local maximum or saddle point. This is done with the Hessian from

$$ (f\circ g_1)(x) = f(x_1,x_2,s-x_1-x_2) $$

NOTE

For $s = 2$ we have

$$ \left( \begin{array}{ccccccccccc} x_1 & x_ 2 & x_3 & \lambda_1 & \lambda_2 & \lambda_3 & \lambda_4 & \epsilon_1 & \epsilon_2 & \epsilon_3 & f(x)\\ 0 & 0 & 2. & 0 & -6. & -1. & 0 & 0 & 0 & -1.41421 & 0. \\ 0 & 0 & 2. & 0 & -6. & -1. & 0 & 0 & 0 & 1.41421 & 0. \\ 0 & 2. & 0 & -1. & -3. & 0 & 1. & 0 & -1.41421 & 0 & 2. \\ 0 & 2. & 0 & -1. & -3. & 0 & 1. & 0 & 1.41421 & 0 & 2. \\ 0.75 & 1.25 & 0 & -2.5 & 0 & 0 & -0.6875 & -0.866025 & -1.11803 & 0 & 3.125 \\ 0.75 & 1.25 & 0 & -2.5 & 0 & 0 & -0.6875 & -0.866025 & 1.11803 & 0 & 3.125 \\ 0.75 & 1.25 & 0 & -2.5 & 0 & 0 & -0.6875 & 0.866025 & -1.11803 & 0 & 3.125 \\ 0.75 & 1.25 & 0 & -2.5 & 0 & 0 & -0.6875 & 0.866025 & 1.11803 & 0 & 3.125 \\ 1. & 0 & 1. & -3. & 0 & -1. & 0 & -1. & 0 & -1. & 3. \\ 1. & 0 & 1. & -3. & 0 & -1. & 0 & -1. & 0 & 1. & 3. \\ 1. & 0 & 1. & -3. & 0 & -1. & 0 & 1. & 0 & -1. & 3. \\ 1. & 0 & 1. & -3. & 0 & -1. & 0 & 1. & 0 & 1. & 3. \\ 2. & 0 & 0 & 0 & 0 & -5. & -6. & -1.41421 & 0 & 0 & 0. \\ 2. & 0 & 0 & 0 & 0 & -5. & -6. & 1.41421 & 0 & 0 & 0. \\ \end{array} \right) $$

The superabundant solutions are due to $\epsilon_k^2$

A solution with $\epsilon_k = 0$ means that the restriction $g_{k+1}(x,\epsilon)$ is active.