Math Problem Statement

Solve part c.

Determine the optimum solution using the Simplex method.

Solution

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, M1M_1 and M2M_2. The table provides the resource constraints and the profit:

Paint TypeRaw Material M1M_1 (tons)Raw Material M2M_2 (tons)Profit per ton ($1000)
Exterior Paint615
Interior Paint424

Resource Constraints:

  • Maximum daily availability of raw material M1M_1 = 24 tons
  • Maximum daily availability of raw material M2M_2 = 6 tons

Let:

  • x1x_1 = tons of exterior paint produced per day
  • x2x_2 = tons of interior paint produced per day

Objective Function:

The objective is to maximize the daily profit: Maximize z=5x1+4x2\text{Maximize } z = 5x_1 + 4x_2

Constraints:

The production is subject to the availability of raw materials:

  1. For M1M_1: 6x1+4x2246x_1 + 4x_2 \leq 24
  2. For M2M_2: 1x1+2x261x_1 + 2x_2 \leq 6
  3. Non-negativity constraints: x10,x20x_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: 6x1+4x2+s1=246x_1 + 4x_2 + s_1 = 24 x1+2x2+s2=6x_1 + 2x_2 + s_2 = 6 where s1s_1 and s2s_2 are slack variables representing unused resources for M1M_1 and M2M_2, respectively. The objective function remains the same: Maximize z=5x1+4x2\text{Maximize } z = 5x_1 + 4x_2

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

\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.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Linear Programming
Optimization
Simplex Method

Formulas

Objective Function: Maximize z = 5x1 + 4x2
Constraint 1 (Material M1): 6x1 + 4x2 <= 24
Constraint 2 (Material M2): 1x1 + 2x2 <= 6

Theorems

Simplex Method
Slack Variables
Pivot Operations

Suitable Grade Level

University Level (Operations Research, Linear Programming)