Multiplying Mixed Numbers Worksheet - Free Printable

Problem Statement:

A company produces two types of products, Product A and Product B, using two machines, Machine X and Machine Y. The production times for each product on each machine are as follows:

- Product A:
- Machine X: 2 hours
- Machine Y: 1 hour

- Product B:
- Machine X: 1 hour
- Machine Y: 3 hours

The company has the following constraints:
1. Machine X is available for a maximum of 8 hours per day.
2. Machine Y is available for a maximum of 9 hours per day.

The profit per unit for Product A is $50, and for Product B is $60.

The company wants to determine how many units of each product to produce daily to maximize their profit.

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Solution Approach:

This is a classic linear programming problem. We will solve it step by step:

#### Step 1: Define the 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 objective is to maximize the total profit. The profit function is given by:
\[
\text{Profit} = 50x + 60y
\]

#### Step 3: Formulate the Constraints
The constraints are based on the availability of Machine X and Machine Y:

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

2. Machine Y constraint:
- Product A requires 1 hour on Machine Y per unit.
- Product B requires 3 hours on Machine Y per unit.
- Total available time on Machine Y is 9 hours.
\[
x + 3y \leq 9
\]

3. 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 linear programming problem can be summarized as:
\[
\text{Maximize } P = 50x + 60y
\]
subject to:
\[
2x + y \leq 8
\]
\[
x + 3y \leq 9
\]
\[
x \geq 0, \quad y \geq 0
\]

#### Step 5: Solve Graphically
To solve this graphically, we plot the constraints and find the feasible region. Then, we evaluate the objective function at the vertices of the feasible region.

##### Plotting 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: \( x + 3y \leq 9 \)
- When \( x = 0 \): \( y = 3 \)
- When \( y = 0 \): \( x = 9 \)
- Line: \( x + 3y = 9 \)

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

##### Feasible Region:
The feasible region is the area where all constraints are satisfied. It is a polygon defined by the intersection of the lines:
- \( 2x + y = 8 \)
- \( x + 3y = 9 \)
- \( x = 0 \)
- \( y = 0 \)

##### Vertices of the Feasible Region:
The vertices of the feasible region are the points where the constraint lines intersect. We find these points by solving the system of equations:

1. Intersection of \( 2x + y = 8 \) and \( x + 3y = 9 \):
\[
\begin{aligned}
&2x + y = 8 \quad \text{(1)} \\
&x + 3y = 9 \quad \text{(2)}
\end{aligned}
\]
From equation (1): \( y = 8 - 2x \).
Substitute into equation (2):
\[
x + 3(8 - 2x) = 9
\]
\[
x + 24 - 6x = 9
\]
\[
-5x + 24 = 9
\]
\[
-5x = -15
\]
\[
x = 3
\]
Substitute \( x = 3 \) back into \( y = 8 - 2x \):
\[
y = 8 - 2(3) = 2
\]
So, the intersection point is \( (3, 2) \).

2. Intersection of \( 2x + y = 8 \) and \( y = 0 \):
\[
2x + 0 = 8 \implies x = 4
\]
So, the intersection point is \( (4, 0) \).

3. Intersection of \( x + 3y = 9 \) and \( x = 0 \):
\[
0 + 3y = 9 \implies y = 3
\]
So, the intersection point is \( (0, 3) \).

4. Intersection of \( x = 0 \) and \( y = 0 \):
The origin \( (0, 0) \).

##### Vertices of the Feasible Region:
The vertices are:
- \( (0, 0) \)
- \( (4, 0) \)
- \( (0, 3) \)
- \( (3, 2) \)

##### Evaluate the Objective Function at Each Vertex:
1. At \( (0, 0) \):
\[
P = 50(0) + 60(0) = 0
\]

2. At \( (4, 0) \):
\[
P = 50(4) + 60(0) = 200
\]

3. At \( (0, 3) \):
\[
P = 50(0) + 60(3) = 180
\]

4. At \( (3, 2) \):
\[
P = 50(3) + 60(2) = 150 + 120 = 270
\]

#### Step 6: Determine the Optimal Solution
The maximum profit occurs at the vertex \( (3, 2) \), where the profit is $270.

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Final Answer:

\[
\boxed{(3, 2)}
\]
The company should produce 3 units of Product A and 2 units of Product B daily to maximize their profit, which will be $270.