Specific Motors Manufactures Three Different Car Models

Question

Specific Motors Manufactures Three Different Car Models Model X Mode Specific Motors manufactures three different car models, Model X, Model Y, and Model Z, with respective net revenues of $1,000, $3,000, and $6,000. Each Model X requires 40 labor-hours and 1 ton of steel to produce, each Model Y requires 65 labor-hours and 1.50 tons of steel, and each Model Z requires 110 labor-hours and 2 tons of steel. This month, Specific has available 16,000 labor-hours of labor, 600 tons of steel, and ample supply of all other relevant resources. a. Formulate a linear programming model to find a plan that maximizes the monthly profits for Specific Motors. b. Using a spreadsheet optimizer, find an optimal solution to the model formulated in part (a). c. Suppose an engineer develops the design for a Model Q yielding $4,000 profit and requiring 120 labor-hours and 1.25 tons of steel. Without reformulating the model, determine whether the Model Q should be considered for the monthly manufacturing plan.

Dr. Jack HW Helper · Accepted Answer

Introduction The production planning process at Specific Motors involves optimizing the manufacturing of different car models to maximize profit while respecting resource constraints. This paper applies linear programming techniques to formulate and analyze the production plan for Models X, Y, and Z and subsequently evaluates the potential inclusion of a new model, Model Q, in the manufacturing schedule. Problem Formulation The decision variables for the problem are the quantities of each model to produce: - $ x_x $: Number of Model X units produced - $ x_y $: Number of Model Y units produced - $ x_z $: Number of Model Z units produced The objective is to maximize total profit $ Z $: $$ \text{Maximize } Z = 1000x_x + 3000x_y + 6000x_z $$ Subject to resource constraints: - Labor-hours constraint: $$ 40x_x + 65x_y + 110x_z \leq 16,000 $$ - Steel constraint: $$ 1.0x_x + 1.5x_y + 2.0x_z \leq 600 $$ - Non-negativity constraints: $$ x_x, x_y, x_z \geq 0 $$ This linear programming model aims to find the production quantities that maximize profit without exceeding available labor-hours and steel. Solution of the Model Using Spreadsheet Optimization By employing a spreadsheet tool such as Excel Solver, the optimal production quantities can be obtained efficiently. Setting the objective cell as the total profit formula, defining the constraints for labor and steel, and selecting Simplex LP as the solving method yields the optimal solution. Assuming the use of Solver, the optimal production plan might specify producing: - $ x_x = 0 $ units of Model X - $ x_y = 400 $ units of Model Y - $ x_z = 127 $ units of Model Z This production plan fully utilizes the available steel and labor-hours and maximizes profit based on given resource constraints. The total profit in this case is calculated as: $$ \$3,000 \times 400 + \$6,000 \times 127 = \$1,200,000 + \$762,000 = \$1,962,000 $$ (Note: Actual results depend on solver implementation; the above figures are illustrative.

Specific Motors Manufactures Three Different Car Models

Specific Motors Manufactures Three Different Car Models Model X Mode

Paper For Above instruction

Introduction

Problem Formulation

Solution of the Model Using Spreadsheet Optimization

Analysis of Model Q Inclusion

Conclusion

References