Minimization in a cold drink company, involving three variables, solved by simplex techniques.

To minimize costs in a cold drink company, we can consider three variables: the cost of ingredients, the cost of packaging, and the cost of production. We can use simplex techniques to optimize these variables and minimize overall costs.

Let's define the variables:

- x1: cost of ingredients per unit of drink
- x2: cost of packaging per unit of drink
- x3: cost of production per unit of drink

The objective function to minimize overall costs can be represented as:

Minimize Z = x1 + x2 + x3

Subject to the following constraints:

- 2x1 + 3x2 + 4x3 ≥ 100 (constraint on total cost)
- x1 + x2 + x3 ≥ 20 (constraint on minimum cost per unit)
- x1, x2, x3 ≥ 0 (non-negativity constraint)

We can now set up the initial simplex table:

| Basic Variables | x1 | x2 | x3 | RHS |
|-----------------|----|----|----|-----|
| x4 | 2 | 3 | 4 | 100 |
| x5 | 1 | 1 | 1 | 20 |
| Z | 1 | 1 | 1 | 0 |

Next, we can perform iterations of the simplex method to find the optimal solution that minimizes overall costs. This involves calculating the pivot column, pivot row, and updating the table accordingly.

After several iterations, we will arrive at the optimal solution that minimizes costs for the cold drink company. This solution will provide the optimal values for x1, x2, and x3 that result in the lowest overall costs while satisfying all constraints.

By using simplex techniques, the cold drink company can efficiently optimize its costs and improve its profitability.

Problem Statement:

ABC Cold Drink Company produces three flavors of cold drinks: Cola, Lemon, and Orange. The company has three production lines and must decide how many gallons of each flavor to produce in order to minimize production costs.

Variables:

- x1 = Gallons of Cola produced
- x2 = Gallons of Lemon produced
- x3 = Gallons of Orange produced

Constraints:

- Production Line 1: 2x1 + x2 + x3 ≤ 600
- Production Line 2: x1 + 2x2 + 2x3 ≤ 400
- Production Line 3: 3x1 + x2 + 2x3 ≤ 500
- Non-negative restrictions: x1, x2, x3 ≥ 0

Objective Function (Minimization):

z = 5x1 + 4x2 + 3x3 (production cost per gallon)

Simplex Tableau:

| Basic Variable | x1 | x2 | x3 | RHS | Ratio |
|---|---|---|---|---|---|
| x4 | 1 | 0 | 0 | 0 | - |
| x5 | 0 | 1 | 0 | 0 | - |
| x6 | 0 | 0 | 1 | 0 | - |
| z | -5 | -4 | -3 | 0 | - |

Iteration 1:

- Entering variable: x1 (smallest negative coefficient in objective row)
- Leaving variable: x4 (smallest ratio of RHS to coefficient of entering variable in column x1)

| Basic Variable | x1 | x2 | x3 | RHS | Ratio |
|---|---|---|---|---|---|
| x1 | 1 | 0 | 0 | 0 | - |
| x5 | 0 | 1 | 0 | 0 | - |
| x6 | 0 | 0 | 1 | 0 | - |
| z | -10 | -4 | -3 | 0 | - |

Iteration 2:

- Entering variable: x3 (smallest negative coefficient in objective row)
- Leaving variable: x6 (smallest ratio of RHS to coefficient of entering variable in column x3)

| Basic Variable | x1 | x2 | x3 | RHS | Ratio |
|---|---|---|---|---|---|
| x1 | 1 | 0 | 0 | 0 | - |
| x5 | 0 | 1 | 0 | 0 | - |
| x3 | 0 | 0 | 1 | 0 | - |
| z | -15 | -4 | 0 | 0 | - |

Iteration 3:

- Entering variable: x2 (smallest negative coefficient in objective row)
- Leaving variable: x5 (smallest ratio of RHS to coefficient of entering variable in column x2)

| Basic Variable | x1 | x2 | x3 | RHS | Ratio |
|---|---|---|---|---|---|
| x1 | 1 | 0 | 0 | 0 | - |
| x2 | 0 | 1 | 0 | 0 | - |
| x3 | 0 | 0 | 1 | 0 | - |
| z | -20 | 0 | 0 | 0 | - |

Optimal Solution:

x1 = 0 gallons of Cola
x2 = 0 gallons of Lemon
x3 = 0 gallons of Orange

Minimum Production Cost:

z = 0

Interpretation:

The simplex technique reveals that the ABC Cold Drink Company should not produce any gallons of any flavor to minimize production costs. This result is likely due to the input constraints, which indicate limited production capacity.