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Problem Statement:

A company produces two types of products, Product A and Product B, using three machines: Machine 1, Machine 2, and Machine 3. The production times (in hours) required for each product on each machine are given in the following table:

| Machine/Product | Product A | Product B |
|-----------------|-----------|-----------|
| Machine 1 | 2 | 1 |
| Machine 2 | 1 | 2 |
| Machine 3 | 3 | 2 |

The company has the following constraints:
1. Machine 1 can operate for a maximum of 8 hours per day.
2. Machine 2 can operate for a maximum of 6 hours per day.
3. Machine 3 can operate for a maximum of 10 hours per day.

The profit per unit for Product A is $5, and for Product B is $4. The company wants to determine how many units of each product to produce daily to maximize its profit.

#### Objective:
Formulate this problem as a linear programming problem and solve it to find the optimal number of units of Product A and Product B to produce daily to maximize profit.

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Step 1: Define Decision Variables

Let:
- \( x \): Number of units of Product A produced daily.
- \( y \): Number of units of Product B produced daily.

---

Step 2: Formulate the Objective Function

The profit from producing \( x \) units of Product A and \( y \) units of Product B is given by:
\[
\text{Profit} = 5x + 4y
\]
The objective is to maximize this profit:
\[
\text{Maximize } Z = 5x + 4y
\]

---

Step 3: Formulate the Constraints

The production times for each machine impose constraints on the values of \( x \) and \( y \):

1. Machine 1 constraint:
- Product A requires 2 hours per unit.
- Product B requires 1 hour per unit.
- Total available time for Machine 1 is 8 hours.
\[
2x + y \leq 8
\]

2. Machine 2 constraint:
- Product A requires 1 hour per unit.
- Product B requires 2 hours per unit.
- Total available time for Machine 2 is 6 hours.
\[
x + 2y \leq 6
\]

3. Machine 3 constraint:
- Product A requires 3 hours per unit.
- Product B requires 2 hours per unit.
- Total available time for Machine 3 is 10 hours.
\[
3x + 2y \leq 10
\]

4. Non-negativity constraints:
- The number of units produced cannot be negative.
\[
x \geq 0, \quad y \geq 0
\]

---

Step 4: Write the Linear Programming Problem

The complete linear programming problem is:
\[
\text{Maximize } Z = 5x + 4y
\]
subject to:
\[
\begin{aligned}
2x + y & \leq 8 \\
x + 2y & \leq 6 \\
3x + 2y & \leq 10 \\
x & \geq 0 \\
y & \geq 0
\end{aligned}
\]

---

Step 5: Solve the Linear Programming Problem

To solve this problem, we will use the graphical method. We will plot the feasible region defined by the constraints and then find the vertices of the feasible region. Finally, we will evaluate the objective function at each vertex to determine the maximum profit.

#### Step 5.1: Plot the Constraints
1. Constraint 1: \( 2x + y \leq 8 \)
- When \( x = 0 \): \( y = 8 \)
- When \( y = 0 \): \( x = 4 \)
- Line: \( 2x + y = 8 \)

2. Constraint 2: \( x + 2y \leq 6 \)
- When \( x = 0 \): \( y = 3 \)
- When \( y = 0 \): \( x = 6 \)
- Line: \( x + 2y = 6 \)

3. Constraint 3: \( 3x + 2y \leq 10 \)
- When \( x = 0 \): \( y = 5 \)
- When \( y = 0 \): \( x = \frac{10}{3} \approx 3.33 \)
- Line: \( 3x + 2y = 10 \)

4. Non-negativity constraints: \( x \geq 0 \) and \( y \geq 0 \)

#### Step 5.2: Find the Feasible Region
The feasible region is the intersection of all the constraints. By plotting these lines, we identify the vertices of the feasible region. The vertices are the points where the constraint lines intersect.

#### Step 5.3: Find the Intersection Points
1. Intersection of \( 2x + y = 8 \) and \( x + 2y = 6 \):
\[
\begin{aligned}
2x + y &= 8 \quad \text{(1)} \\
x + 2y &= 6 \quad \text{(2)}
\end{aligned}
\]
Multiply equation (2) by 2:
\[
2x + 4y = 12 \quad \text{(3)}
\]
Subtract equation (1) from equation (3):
\[
(2x + 4y) - (2x + y) = 12 - 8 \\
3y = 4 \\
y = \frac{4}{3}
\]
Substitute \( y = \frac{4}{3} \) into equation (2):
\[
x + 2\left(\frac{4}{3}\right) = 6 \\
x + \frac{8}{3} = 6 \\
x = 6 - \frac{8}{3} = \frac{18}{3} - \frac{8}{3} = \frac{10}{3}
\]
Intersection point: \( \left( \frac{10}{3}, \frac{4}{3} \right) \)

2. Intersection of \( 2x + y = 8 \) and \( 3x + 2y = 10 \):
\[
\begin{aligned}
2x + y &= 8 \quad \text{(1)} \\
3x + 2y &= 10 \quad \text{(2)}
\end{aligned}
\]
Multiply equation (1) by 2:
\[
4x + 2y = 16 \quad \text{(3)}
\]
Subtract equation (2) from equation (3):
\[
(4x + 2y) - (3x + 2y) = 16 - 10 \\
x = 6
\]
Substitute \( x = 2 \) into equation (1):
\[
2(2) + y = 8 \\
4 + y = 8 \\
y = 4
\]
Intersection point: \( (2, 4) \)

3. Intersection of \( x + 2y = 6 \) and \( 3x + 2y = 10 \):
\[
\begin{aligned}
x + 2y &= 6 \quad \text{(1)} \\
3x + 2y &= 10 \quad \text{(2)}
\end{aligned}
\]
Subtract equation (1) from equation (2):
\[
(3x + 2y) - (x + 2y) = 10 - 6 \\
2x = 4 \\
x = 2
\]
Substitute \( x = 2 \) into equation (1):
\[
2 + 2y = 6 \\
2y = 4 \\
y = 2
\]
Intersection point: \( (2, 2) \)

4. Intersection with axes:
- When \( x = 0 \) in \( 2x + y = 8 \): \( y = 8 \). Point: \( (0, 8) \)
- When \( y = 0 \) in \( x + 2y = 6 \): \( x = 6 \). Point: \( (6, 0) \)
- When \( x = 0 \) in \( 3x + 2y = 10 \): \( y = 5 \). Point: \( (0, 5) \)
- When \( y = 0 \) in \( 3x + 2y = 10 \): \( x = \frac{10}{3} \). Point: \( \left( \frac{10}{3}, 0 \right) \)

#### Step 5.4: Identify the Feasible Region Vertices
The feasible region is a polygon with vertices:
\[
(0, 0), \quad (0, 3), \quad (2, 2), \quad \left( \frac{10}{3}, \frac{4}{3} \right), \quad (2, 4), \quad (0, 5)
\]

#### Step 5.5: Evaluate the Objective Function at Each Vertex
1. At \( (0, 0) \):
\[
Z = 5(0) + 4(0) = 0
\]

2. At \( (0, 3) \):
\[
Z = 5(0) + 4(3) = 12
\]

3. At \( (2, 2) \):
\[
Z = 5(2) + 4(2) = 10 + 8 = 18
\]

4. At \( \left( \frac{10}{3}, \frac{4}{3} \right) \):
\[
Z = 5\left( \frac{10}{3} \right) + 4\left( \frac{4}{3} \right) = \frac{50}{3} + \frac{16}{3} = \frac{66}{3} = 22
\]

5. At \( (2, 4) \):
\[
Z = 5(2) + 4(4) = 10 + 16 = 26
\]

6. At \( (0, 5) \):
\[
Z = 5(0) + 4(5) = 20
\]

#### Step 5.6: Determine the Maximum Profit
The maximum value of \( Z \) is 26, which occurs at the point \( (2, 4) \).

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Final Answer:

The company should produce 2 units of Product A and 4 units of Product B daily to maximize its profit. The maximum profit is:
\[
\boxed{26}
\]