Worksheet on Significant Figures - 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 10 hours per day.
2. Machine Y is available for a maximum of 12 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.

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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:
\[
\text{Profit} = 5x + 4y
\]

#### 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 10 hours.
\[
2x + y \leq 10
\]

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 12 hours.
\[
x + 3y \leq 12
\]

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 now be written as:
\[
\text{Maximize } P = 5x + 4y
\]
subject to:
\[
2x + y \leq 10
\]
\[
x + 3y \leq 12
\]
\[
x \geq 0, \quad y \geq 0
\]

#### 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: \( 2x + y \leq 10 \)
- When \( x = 0 \): \( y = 10 \)
- When \( y = 0 \): \( x = 5 \)
- Line: \( 2x + y = 10 \)

2. Constraint 2: \( x + 3y \leq 12 \)
- When \( x = 0 \): \( y = 4 \)
- When \( y = 0 \): \( x = 12 \)
- Line: \( x + 3y = 12 \)

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:
- \( 2x + y = 10 \)
- \( x + 3y = 12 \)
- \( x = 0 \)
- \( y = 0 \)

##### 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 formed by pairs of constraints.

1. Intersection of \( 2x + y = 10 \) and \( x + 3y = 12 \):
\[
\begin{aligned}
&2x + y = 10 \quad \text{(1)} \\
&x + 3y = 12 \quad \text{(2)}
\end{aligned}
\]
From (1): \( y = 10 - 2x \).
Substitute into (2):
\[
x + 3(10 - 2x) = 12
\]
\[
x + 30 - 6x = 12
\]
\[
-5x + 30 = 12
\]
\[
-5x = -18
\]
\[
x = \frac{18}{5} = 3.6
\]
Substitute \( x = 3.6 \) back into \( y = 10 - 2x \):
\[
y = 10 - 2(3.6) = 10 - 7.2 = 2.8
\]
Vertex: \( (3.6, 2.8) \)

2. Intersection of \( 2x + y = 10 \) and \( y = 0 \):
\[
2x + 0 = 10 \implies x = 5
\]
Vertex: \( (5, 0) \)

3. Intersection of \( x + 3y = 12 \) and \( x = 0 \):
\[
0 + 3y = 12 \implies y = 4
\]
Vertex: \( (0, 4) \)

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

##### Evaluate the Objective Function at Each Vertex:
The objective function is \( P = 5x + 4y \). We evaluate it at each vertex:

1. At \( (3.6, 2.8) \):
\[
P = 5(3.6) + 4(2.8) = 18 + 11.2 = 29.2
\]

2. At \( (5, 0) \):
\[
P = 5(5) + 4(0) = 25
\]

3. At \( (0, 4) \):
\[
P = 5(0) + 4(4) = 16
\]

4. At \( (0, 0) \):
\[
P = 5(0) + 4(0) = 0
\]

#### Step 6: Determine the Optimal Solution
The maximum profit occurs at the vertex \( (3.6, 2.8) \) with a profit of \( 29.2 \).

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Final Answer:

\[
\boxed{(3.6, 2.8)}
\]
The company should produce 3.6 units of Product A and 2.8 units of Product B daily to maximize its profit, which will be \$29.2. If fractional production is not allowed, the company may need to round down to the nearest whole numbers and re-evaluate the profit. However, the exact solution based on the given constraints is \( (3.6, 2.8) \).