Math Problem Statement
A factory produces two types of products, A and B. Each unit of product A yields a profit of $5, while each unit of product B yields a profit of $3. The factory has limited resources and can produce up to: 4 hours of machine time available for production. 12 units of raw material available. No more than 6 units of product A can be produced. The objective is to maximize profit, subject to these constraints.
Let: x: Number of units of product A produced y: Number of units of product B produced Linear Function (Profit to maximize): Maximize Z = 5 x + 3y Subject to the constraints: Machine time constraint: x + y ≤ 4 Raw material constraint: 2x + y ≤ 12 Production limit for product A: x ≤ 6 Non-negativity constraints: x ≥ 0, y ≥ 0
Solution
To solve this linear programming problem, we will maximize the profit function , subject to the given constraints:
Constraints:
- (Machine time constraint)
- (Raw material constraint)
- (Production limit for product A)
- , (Non-negativity constraints)
Steps to solve:
- Identify the feasible region determined by the constraints.
- Find the vertices of the feasible region (corner points).
- Evaluate the profit function at each vertex.
- Select the vertex that yields the maximum .
Let's find the vertices of the feasible region by solving the constraints pairwise.import numpy as np import pandas as pd from scipy.optimize import linprog
Coefficients of the profit function (objective function)
c = [-5, -3] # Coefficients for maximizing (negative for minimization in linprog)
Coefficients of inequality constraints (Ax <= b)
A = [[1, 1], # Machine time constraint [2, 1], # Raw material constraint [1, 0]] # Production limit for product A b = [4, # Machine time availability 12, # Raw material availability 6] # Maximum production limit for A
Bounds for x and y (non-negativity and limits)
x_bounds = (0, None) # x >= 0 y_bounds = (0, None) # y >= 0
Solve the linear programming problem
result = linprog(c, A_ub=A, b_ub=b, bounds=[x_bounds, y_bounds], method='highs')
Extract the results
x_opt, y_opt = result.x # Optimal values of x and y max_profit = -result.fun # Maximum profit (negating because we minimized)
Display the results in a user-friendly format
optimal_solution = { "Variable": ["x (Product A)", "y (Product B)", "Profit"], "Optimal Value": [x_opt, y_opt, max_profit] }
optimal_df = pd.DataFrame(optimal_solution) import ace_tools as tools; tools.display_dataframe_to_user(name="Optimal Production Solution", dataframe=optimal_df) Variable Optimal Value 0 x (Product A) 4.0 1 y (Product B) 0.0 2 Profit 20.0 Variable Optimal Value 0 x (Product A) 4.0 1 y (Product B) 0.0 2 Profit 20.0successThe optimal production solution has been calculated and displayed in a table for you. You can review the number of units of Product A and Product B to produce, along with the maximum achievable profit.
Follow-Up Questions:
- What would happen to the profit if more machine time was available?
- How would the solution change if the profit per unit of Product B increased?
- What are the implications if the raw material limit is relaxed?
- Can the factory produce both products and still maximize profit under the current constraints?
- How sensitive is the optimal solution to changes in production limits?
Tip:
Understanding the shadow price of constraints can help determine which resource is most valuable and where investing additional resources will yield the highest profit.
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Math Problem Analysis
Mathematical Concepts
Linear Programming
Optimization
Inequalities
Feasible Region
Formulas
Objective Function: Z = 5x + 3y
Constraints: x + y ≤ 4, 2x + y ≤ 12, x ≤ 6, x ≥ 0, y ≥ 0
Theorems
Fundamental Theorem of Linear Programming
Suitable Grade Level
Grades 10-12
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