A company has two machines for manufacturing a product, Machine 1 makes 2 units per hour, while machine 2 makes 3 units per hour. The company has an order of 80 units. Energy restrictions dictate that only one machine can operate at one time. The company has 40 hours of regular machine time, but overtime is available. It costs $4 to run machine for 1 hour, while machine 2 costs $5 per hour. As for the goals:
Meet the demand of 80 units.

To determine the optimal way to meet the demand of 80 units given the constraints and costs, we need to solve a linear programming problem. Here's a step-by-step outline:

### Define the Variables:

- Let $ x_1 $ be the number of hours Machine 1 operates.

- Let $ x_2 $ be the number of hours Machine 2 operates.

### Objective Function:

Minimize the total cost:

\[ \text{Minimize Cost} = 4x_1 + 5x_2 \]

### Constraints:

1. **Demand Constraint**: The total units produced should be at least 80 units.

\[ 2x_1 + 3x_2 \geq 80 \]

2. **Regular Time Constraint**: The total hours should not exceed 40 hours.

\[ x_1 + x_2 \leq 40 \]

3. **Non-negativity Constraint**:

\[ x_1 \geq 0 \]

\[ x_2 \geq 0 \]

### Steps to Solve the Linear Programming Problem:

1. **Formulate the Problem**: Write down the objective function and constraints clearly.

2. **Graph the Constraints**: Plot the constraints on a graph to find the feasible region.

3. **Find the Intersection Points**: Determine where the constraint lines intersect.

4. **Evaluate the Objective Function at Each Vertex of the Feasible Region**: The minimum cost will occur at one of these vertices.

Let's proceed with these steps.

### Graphing the Constraints:

The two main constraints are:

1. $ 2x_1 + 3x_2 \geq 80 $

2. $ x_1 + x_2 \leq 40 $

First, we need to find the intersection points by solving these equations simultaneously.

### Solving for Intersections:

1. Solve for the point where $ 2x_1 + 3x_2 = 80 $ intersects $ x_1 + x_2 = 40 $:

\[ x_1 + x_2 = 40 \]

\[ 2x_1 + 3x_2 = 80 \]

Multiply the first equation by 2:

\[ 2x_1 + 2x_2 = 80 \]

Subtract this from the second equation:

\[ (2x_1 + 3x_2) - (2x_1 + 2x_2) = 80 - 80 \]

\[ x_2 = 0 \]

Substitute $ x_2 = 0 $ into $ x_1 + x_2 = 40 $:

\[ x_1 + 0 = 40 \]

\[ x_1 = 40 \]

So, one intersection point is (40, 0).

2. Find the x-intercept of $ 2x_1 + 3x_2 = 80 $:

Set $ x_2 = 0 $:

\[ 2x_1 = 80 \]

\[ x_1 = 40 \]

3. Find the y-intercept of $ 2x_1 + 3x_2 = 80 $:

Set $ x_1 = 0 $:

\[ 3x_2 = 80 \]

\[ x_2 = \frac{80}{3} \approx 26.67 \]

4. Intersection of $ x_1 + x_2 = 40 $ with the x and y axes:

\[ x_1 = 0 \Rightarrow x_2 = 40 \]

\[ x_2 = 0 \Rightarrow x_1 = 40 \]

Now we can evaluate the objective function at these points:

- (40, 0)

- (0, 40)

- (0, 26.67)

### Evaluate the Objective Function at Each Point:

1. At (40, 0):

\[ 4x_1 + 5x_2 = 4(40) + 5(0) = 160 \]

2. At (0, 40):

\[ 4x_1 + 5x_2 = 4(0) + 5(40) = 200 \]

3. At (0, 26.67):

\[ 4x_1 + 5x_2 = 4(0) + 5(26.67) = 133.35 \]

The minimum cost occurs at (0, 26.67) with a cost of 133.35.

### Conclusion:

To meet the demand of 80 units at the minimum cost:

- Run Machine 1 for 0 hours.

- Run Machine 2 for approximately 26.67 hours.

The minimum cost will be $133.35.