Each day, the production manager is faced with the problem of how many of two types of products to produce. The manager’s goal is to maximize profits, but, at the same time, no existing customer orders should go unfilled. Based on the following information, how many of each product should the manager produce today?

Product	Aluminum required (pounds)	Steel required (pounds)	Profit per unit	Customer orders
A	6	3	7	5
B	5	3	12	3
Current inventory of aluminum – 72 pounds

Current inventory of steel – 93 pounds

Note: Non-integer quantities for both products are allowed. Round your answer to one decimal place.



To solve this optimization problem, we’ll set up a linear programming model.

### Step 1: Define the Variables
Let:
- \( x \) = quantity of product A to produce.
- \( y \) = quantity of product B to produce.

### Step 2: Set Up the Objective Function
The goal is to **maximize the profit**, which is given by:
\[
\text{Profit} = 7x + 12y
\]

### Step 3: Set Up the Constraints

1. **Material Constraints (Aluminum and Steel)**:
   - Aluminum: Each unit of A requires 6 pounds, and each unit of B requires 5 pounds. The total aluminum used should not exceed 72 pounds:
     \[
     6x + 5y \leq 72
     \]
   - Steel: Each unit of A requires 3 pounds, and each unit of B requires 3 pounds. The total steel used should not exceed 93 pounds:
     \[
     3x + 3y \leq 93
     \]

2. **Demand Constraints**:
   - The production of A should at least meet the customer orders for A (5 units):
     \[
     x \geq 5
     \]
   - The production of B should at least meet the customer orders for B (3 units):
     \[
     y \geq 3
     \]

3. **Non-negativity**:
   - \( x \geq 0 \)
   - \( y \geq 0 \)

### Step 4: Solve the Linear Program
The linear programming problem is as follows:

**Maximize**:
\[
7x + 12y
\]

**Subject to**:
\[
6x + 5y \leq 72
\]
\[
3x + 3y \leq 93
\]
\[
x \geq 5
\]
\[
y \geq 3
\]

Let’s solve this using an optimization tool.The optimal solution is to produce:

- **5 units of product A**
- **8.4 units of product B**

This will result in a **maximum profit of 135.8**.

Would you like more details or have any questions?

Here are 5 related questions for further exploration:

1. How would a change in aluminum or steel inventory affect production decisions?
2. What if customer demand constraints were relaxed—would profit increase?
3. How could this model be adapted if only integer quantities were allowed?
4. What impact does an increase in profit per unit of product B have on production?
5. How would additional products impact the formulation of this linear program?

**Tip:** In linear programming, always confirm constraints are realistic to ensure practical solutions.

Each day, the production manager is faced with the problem of how many of two types of products to produce. The manager’s goal is to maximize profits, but, at the same time, no existing customer orders should go unfilled. Based on the following information, how many of each product should the manager produce today?

Product    Aluminum required (pounds)    Steel required (pounds)    Profit per unit    Customer orders
A    6    3    7    5
B    5    3    12    3

Current inventory of aluminum – 72 pounds
Current inventory of steel – 93 pounds

Note: Non-integer quantities for both products are allowed. Round your answer to one decimal place.

This problem explores how a production manager can determine the optimal quantity of products A and B to maximize profits while meeting customer demands and material constraints. Using linear programming, students will apply optimization techniques to maximize profit given constraints on aluminum, steel, and minimum production levels. This is suitable for students in Grades 11-12.

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