Let's analyze the problem step by step: --- ## **Problem Breakdown** - **Total budget for new equipment:** \$4,000,000 - **Models:** - **F300** - Cost per unit: \$350,000 - Annual maintenance per unit: \$12,000 - Units produced per day: 125 - **S250** - Cost per unit: \$220,000 - Annual maintenance per unit: \$7,500 - Units produced per day: 81 - **Additional info:** - At least 1/3 of the models purchased should be F300 - The number of units produced per day is limited by the number of units purchased - The goal is to maximize daily production --- ## **Step 1: Define Variables** Let: - \( x \) = number of F300 units purchased - \( y \) = number of S250 units purchased --- ## **Step 2: Write the Cost Constraint** Total cost: \[ 350,000x + 220,000y \leq 4,000,000 \] --- ## **Step 3: Set the production constraints** - Daily units: \[ \text{F300 units per day} = 125x \] - S250 units per day: \[ 81y \] - Total daily production: \[ P = 125x + 81y \] --- ## **Step 4: Set the constraint for the ratio of F300 to total** Because at least 1/3 of the total models should be F300: \[ x \geq \frac{1}{3}(x + y) \implies 3x \geq x + y \implies 2x \geq y \] Or: \[ y \leq 2x \] --- ## **Step 5: Non-negativity constraints** \[ x \geq , \quad y \geq \] --- ## **Step 6: Objective Function** Maximize total daily units: \[ Z = 125x + 81y \] --- ## **Step 7: Summarize the LP problem** \[ \begin{cases} \text{Maximize } Z = 125x + 81y \\ \text{Subject to:} \\ 350,000x + 220,000y \leq 4,000,000 \\ y \leq 2x \\ x \geq , \quad y \geq \end{cases} \] --- ## **Step 8: Find the feasible region** - From the cost constraint: \[ 350,000x + 220,000y \leq 4,000,000 \] Divide through by 10,000: \[ 35x + 22y \leq 400 \] - From \( y \leq 2x \) - Both \( x, y \geq \) --- ## **Step 9: Find the corner points** ### **Point 1:** Intersection with axes: - When \( y= \): \[ 35x \leq 400 \implies x \leq \frac{400}{35} \approx 11.43 \] - When \( x= \): \[ 22y \leq 400 \implies y \leq \frac{400}{22} \approx 18.18 \] ### **Point 2:** Intersection of \( y = 2x \) with cost constraint: Substitute \( y=2x \): \[ 35x + 22(2x) = 400 \] \[ 35x + 44x = 400 \] \[ 79x = 400 \] \[ x = \frac{400}{79} \approx 5.06 \] Then: \[ y = 2x \approx 10.12 \] Check if this point satisfies all constraints: - \( y \leq 2x \): yes, equality - \( x \geq \), \( y \geq \): yes --- ## **Step 10: Calculate the objective at the corner points** ### **Point A:** \( (x, y) = (,) \) \[ Z=125()+81()= \] ### **Point B:** \( (x, y) = (11.43, ) \) \[ Z=125(11.43)+81()=125 \times 11.43 \approx 1,429.38 \] ### **Point C:** \( (x, y) = (5.06, 10.12) \) \[ Z=125 \times 5.06 + 81 \times 10.12 \] \[ = 632.5 + 819.72 \approx 1452.22 \] --- ## **Step 11: Final Decision** The maximum production occurs at \( (x, y) \approx (5.06, 10.12) \), with an approximate daily production of **1452 units**. --- ## **## Final Summary** - Purchase approximately **5 F300 units** and **10 S250 units**. - This combination respects the budget constraint and the ratio constraint. - The maximum daily production achievable is approximately **1452 units**. --- Let me know if you'd like the exact fractional values or further elaboration!

# Step-by-Step Solution Summary ## **Problem Breakdown** United Manufacturing plans to invest up to $4,000,000 in two models of machinery: F300 and S250. Key constraints include budget limits, maintenance costs, production capabilities, and a requirement for at least one-third of the models to be F300. ## **Step 1: Define Variables** Define \( x \) as the number of F300 units and \( y \) as the number of S250 units. This establishes the variables needed for further calculations and constraints. ## **Step 2: Write the Cost Constraint** The total cost constraint is expressed as \( 350,000x + 220,000y \leq 4,000,000 \). This ensures that the total expenditure does not exceed the budget. ## **Step 3: Set the Production Constraints** The production capabilities are defined as \( P = 125x + 81y \). This equation calculates the total daily production based on the number of units purchased. ## **Step 4: Set the Ratio Constraint for F300** The requirement that at least one-third of models purchased should be F300 leads to the constraint \( y \leq 2x \). This ensures a minimum ratio of F300 units to total units. ## **Step 5: Non-negativity Constraints** Non-negativity constraints are defined as \( x \geq 0 \) and \( y \geq 0 \). This stipulates that the number of units purchased cannot be negative. ## **Step 6: Objective Function** The objective function is to maximize daily production: \( Z = 125x + 81y \). This function represents the goal of the optimization problem. ## **Step 7: Summarize the LP Problem** The linear programming problem is summarized with the objective function and all constraints. This layout provides a clear framework for applying optimization techniques. ## **Step 8: Find the Feasible Region** The feasible region is defined by the cost and ratio constraints. Graphing these inequalities will help identify the boundaries of potential solutions. ## **Step 9: Find the Corner Points** Calculate the corner points by finding intersections of the constraints. These points are critical for evaluating the objective function. ## **Step 10: Calculate the Objective at the Corner Points** Evaluate the production function \( Z \) at the identified corner points. This step determines which combination of units yields maximum production. ## **Step 11: Final Decision** The optimal solution is approximately \( x \approx 5.06 \) (F300) and \( y \approx 10.12 \) (S250), resulting in a maximum daily production of approximately **1452 units**. This selection meets all constraints effectively. ## **Final Summary** The strategy involves purchasing about **5 F300 units** and **10 S250 units** to maximize production while adhering to budget and ratio constraints. The calculated daily production capacity with this configuration is approximately **1452 units**.