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


To solve this linear programming problem, we will maximize the profit function $ Z = 5x + 3y $, subject to the given constraints:

### Constraints:
1. $ x + y \leq 4 $ (Machine time constraint)
2. $ 2x + y \leq 12 $ (Raw material constraint)
3. $ x \leq 6 $ (Production limit for product A)
4. $ x \geq 0 $, $ y \geq 0 $ (Non-negativity constraints)

### Steps to solve:
1. Identify the feasible region determined by the constraints.
2. Find the vertices of the feasible region (corner points).
3. Evaluate the profit function $ Z = 5x + 3y $ at each vertex.
4. Select the vertex that yields the maximum $ Z $.

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:
1. What would happen to the profit if more machine time was available?
2. How would the solution change if the profit per unit of Product B increased?
3. What are the implications if the raw material limit is relaxed?
4. Can the factory produce both products and still maximize profit under the current constraints?
5. 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.

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.

Explore how to maximize profits for a factory producing Products A and B using linear programming. This problem involves solving an objective function Z = 5x + 3y subject to constraints like machine time, raw materials, and production limits. Suitable for Grades 10-12 students studying optimization techniques.

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