Answer :
Maximize Profit = 600C + 400D, subject to 24C + 2D ≤ 24, 4C + D ≤ 24, 2C + D ≤ 24, 10(2C + D) ≤ Budget, C ≥ 0, D ≥ 0.
To formulate the linear programming (LP) model, let's define the decision variables and objective function first.
Decision Variables:
Let's define the following decision variables:
- Let C represent the number of times the factory produces 100 kilograms of chocolate.
- Let D represent the number of times the factory produces 100 kilograms of candy.
Objective Function:
The objective is to maximize the profit per day. Since the profit depends on the quantities of chocolate and candy produced, the objective function is as follows:
Maximize: Profit = 600C + 400D
Constraints:
1. Machine A constraint: The available hours for machine A can be represented as 24C + 2D (as 1 hour is required for chocolate and 2 hours for candy for each production).
- Constraint 1: 24C + 2D ≤ 24 (as there are 24 hours available in a day).
2. Machine B constraint: The available hours for machine B can be represented as 4C + D (as 4 hours are required for chocolate and 1 hour for candy for each production).
- Constraint 2: 4C + D ≤ 24 (as there are 24 hours available in a day).
3. Machine C constraint: The available hours for machine C can be represented as 2C + D (as 2 hours are required for chocolate and 1 hour for candy for each production). Since machine C is rented and costs 10 pounds per hour, this cost needs to be considered.
- Constraint 3: 2C + D ≤ 24 (as there are 24 hours available in a day).
- Constraint 4: 10(2C + D) ≤ Budget (to ensure the cost of renting machine C is within the budget).
4. Non-negativity constraints: The number of times the factory produces chocolate and candy cannot be negative.
- Constraint 5: C ≥ 0
- Constraint 6: D ≥ 0
In summary, the LP model can be written as follows:
Maximize: Profit = 600C + 400D
Subject to:
1. 24C + 2D ≤ 24
2. 4C + D ≤ 24
3. 2C + D ≤ 24
4. 10(2C + D) ≤ Budget
5. C ≥ 0
6. D ≥ 0
The objective is to find the values of C and D that maximize the profit while satisfying the constraints. The LP solver can be used to solve this model, providing the optimal values for C and D, and consequently, the maximum profit.
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