Question 1174704
Here's how to solve this linear programming problem using the simplex method:

**1. Define Variables**

* Let x1 be the number of liters of the first drink.
* Let x2 be the number of liters of the second drink.

**2. Formulate the Objective Function**

* The objective is to maximize profit (P).
* Profit = 0.60x1 + 0.50x2
* Maximize P = 0.6x1 + 0.5x2

**3. Formulate the Constraints**

* **Apple Juice Constraint:**
    * 0.3x1 + 0.6x2 ≤ 1000
* **Pineapple Juice Constraint:**
    * 0.7x1 + 0.4x2 ≤ 1500
* **Non-negativity Constraints:**
    * x1 ≥ 0
    * x2 ≥ 0

**4. Set up the Simplex Tableau**

* Introduce slack variables (s1, s2) to convert inequalities to equalities:
    * 0.3x1 + 0.6x2 + s1 = 1000
    * 0.7x1 + 0.4x2 + s2 = 1500
* Rewrite the objective function:
    * -0.6x1 - 0.5x2 + P = 0

The initial simplex tableau is:

| Basic | x1    | x2    | s1    | s2    | Solution |
| :---- | :---- | :---- | :---- | :---- | :------- |
| s1    | 0.3   | 0.6   | 1     | 0     | 1000     |
| s2    | 0.7   | 0.4   | 0     | 1     | 1500     |
| P     | -0.6  | -0.5  | 0     | 0     | 0        |

**5. Perform Simplex Iterations**

* **Choose the Pivot Column:** Select the column with the most negative value in the P row (x1 column).
* **Choose the Pivot Row:** Divide the solution column by the corresponding values in the pivot column and select the row with the smallest non-negative ratio.
    * 1000 / 0.3 ≈ 3333.33
    * 1500 / 0.7 ≈ 2142.86
    * Pivot row is s2.
* **Pivot Element:** 0.7
* **Perform Row Operations:**
    * Divide the pivot row by the pivot element (0.7).
    * Use row operations to make all other elements in the pivot column zero.

After several iterations, you'll reach the optimal solution.

**6. Optimal Solution (from the provided code execution result)**

The python code provided gives the optimal solution.
* x1 ≈ 1666.67 liters (drink 1)
* x2 ≈ 833.33 liters (drink 2)
* Maximum Profit ≈ $1416.67

**Therefore, the company should produce approximately 1666.67 liters of the first drink and 833.33 liters of the second drink to maximize profit, resulting in a profit of approximately $1416.67.**