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To solve part (c), we need to determine the optimum solution using the **Simplex method**. Here’s the problem and step-by-step approach:

### Problem Overview:
Reddy Mikks produces interior and exterior paints using two raw materials, $ M_1 $ and $ M_2 $. The table provides the resource constraints and the profit:

| Paint Type     | Raw Material $ M_1 $ (tons) | Raw Material $ M_2 $ (tons) | Profit per ton ($1000) |
|----------------|-------------------------------|-------------------------------|------------------------|
| Exterior Paint | 6                             | 1                             | 5                      |
| Interior Paint | 4                             | 2                             | 4                      |

#### Resource Constraints:
- Maximum daily availability of raw material $ M_1 $ = 24 tons
- Maximum daily availability of raw material $ M_2 $ = 6 tons

Let:
- $ x_1 $ = tons of exterior paint produced per day
- $ x_2 $ = tons of interior paint produced per day

### Objective Function:
The objective is to **maximize the daily profit**:
\[
\text{Maximize } z = 5x_1 + 4x_2
\]

### Constraints:
The production is subject to the availability of raw materials:
1. For $ M_1 $: $ 6x_1 + 4x_2 \leq 24 $
2. For $ M_2 $: $ 1x_1 + 2x_2 \leq 6 $
3. Non-negativity constraints: $ x_1 \geq 0, x_2 \geq 0 $

### Simplex Method:

1. **Formulation in Standard Form:**
   To apply the Simplex method, we convert inequalities into equalities by adding slack variables:
   \[
   6x_1 + 4x_2 + s_1 = 24
   \]
   \[
   x_1 + 2x_2 + s_2 = 6
   \]
   where $ s_1 $ and $ s_2 $ are slack variables representing unused resources for $ M_1 $ and $ M_2 $, respectively. The objective function remains the same:
   \[
   \text{Maximize } z = 5x_1 + 4x_2
   \]

2. **Initial Simplex Tableau:**
   We can now set up the initial tableau for the Simplex method.

\[
\begin{array}{c|cccccc|c}
\text{Basis} & x_1 & x_2 & s_1 & s_2 & \text{RHS} \\
\hline
s_1 & 6 & 4 & 1 & 0 & 24 \\
s_2 & 1 & 2 & 0 & 1 & 6 \\
\hline
z & -5 & -4 & 0 & 0 & 0 \\
\end{array}
\]

3. **Simplex Iterations:**
   - In the tableau, we see that the most negative coefficient in the objective function is -5 (under $ x_1 $), so $ x_1 $ will enter the basis.
   - We now need to determine which variable will leave the basis by performing the ratio test on the right-hand side (RHS) values. For the ratios:
     - Row 1: $ 24 / 6 = 4 $
     - Row 2: $ 6 / 1 = 6 $
   
   Since the smallest ratio is 4, the pivot element is 6 in the first row. We will now pivot around this element to form the next tableau.

4. **Resulting Tableau and Optimal Solution:**

Let me compute the next steps to find the optimal solution using the Simplex method.The optimal solution using the Simplex method is:

- $ x_1 = 3 $ tons of exterior paint
- $ x_2 = 1.5 $ tons of interior paint

### Maximum Profit:
The maximum daily profit is:
\[
z = 5x_1 + 4x_2 = 5(3) + 4(1.5) = 15 + 6 = 21 \text{ thousand dollars}
\]
Thus, the maximum profit is $21,000 per day.

Would you like further details on the Simplex steps, or have any other questions?

---

### Related Questions:
1. How would the solution change if the profit for exterior paint increases to $6,000 per ton?
2. What would happen if the availability of raw material $ M_2 $ decreases to 5 tons per day?
3. Can you graph the constraints and identify the feasible region for this problem?
4. How does changing the constraints affect the slack variables?
5. What is the dual of this linear programming problem?

### Tip:
Understanding how to read the final Simplex tableau can help you analyze the impact of changes in constraints or objective function coefficients.

Determine the optimum solution using the Simplex method.

Learn how to solve a linear programming problem using the Simplex method. The problem involves maximizing profit for Reddy Mikks paints, with constraints on raw materials and using slack variables for optimization. This step-by-step guide is suitable for university students in Operations Research or related fields.

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