 Solve the LP problem. If no optimal solution exists, indicate whether the feasible region is empty or the objective function is unbounded. (Enter EMPTY if the region is empty. Enter UNBOUNDED if the function is unbounded.)
Minimize c = 2x + 4y subject to
0.3x + 0.3y ≥ 3
3x + 6y ≥ 42
x ≥ 0, y ≥ 0.
Minimum value c = 
 occurs at the endpoints 
(x, y) = 

 
  

 (smaller x-value) and 
(x, y) = 

 
  

 (larger x-value) and 
---Select---
 .

Let's solve this Linear Programming (LP) problem step-by-step.

### Problem
Objective:
\[
\text{Minimize } c = 2x + 4y
\]

Subject to constraints:
1. \( 0.3x + 0.3y \geq 3 \)
2. \( 3x + 6y \geq 42 \)
3. \( x \geq 0 \)
4. \( y \geq 0 \)

### Step 1: Simplify Constraints

1. For the first constraint:
   \[
   0.3x + 0.3y \geq 3
   \]
   Dividing both sides by 0.3, we get:
   \[
   x + y \geq 10
   \]

2. For the second constraint:
   \[
   3x + 6y \geq 42
   \]
   Dividing both sides by 3, we get:
   \[
   x + 2y \geq 14
   \]

### Step 2: Identify Feasible Region

The feasible region is the area that satisfies all constraints. Let's analyze the inequalities to find the points of intersection.

#### Convert the inequalities into equations:

1. \( x + y = 10 \)
2. \( x + 2y = 14 \)

Solving these equations will give us the vertices of the feasible region.

### Step 3: Solve for Intersection Point

1. From \( x + y = 10 \), we get \( y = 10 - x \).
2. Substitute \( y = 10 - x \) into \( x + 2y = 14 \):
   \[
   x + 2(10 - x) = 14
   \]
   \[
   x + 20 - 2x = 14
   \]
   \[
   -x = -6
   \]
   \[
   x = 6
   \]
   Substitute \( x = 6 \) back into \( y = 10 - x \):
   \[
   y = 10 - 6 = 4
   \]

Thus, the intersection point is \( (6, 4) \).

### Step 4: Identify Feasible Vertices

The feasible vertices, given the constraints \( x \geq 0 \) and \( y \geq 0 \), are:
1. The point \( (6, 4) \) from the intersection of the two lines.
2. The point where \( x + y = 10 \) intersects \( y = 0 \): \( x = 10 \), so \( (10, 0) \).
3. The point where \( x + 2y = 14 \) intersects \( x = 0 \): \( y = 7 \), so \( (0, 7) \).

### Step 5: Evaluate the Objective Function at Each Vertex

Now we will calculate \( c = 2x + 4y \) at each vertex to find the minimum value.

1. At \( (6, 4) \):
   \[
   c = 2(6) + 4(4) = 12 + 16 = 28
   \]

2. At \( (10, 0) \):
   \[
   c = 2(10) + 4(0) = 20
   \]

3. At \( (0, 7) \):
   \[
   c = 2(0) + 4(7) = 28
   \]

### Conclusion

The minimum value of \( c \) is **20**, which occurs at the vertex \( (10, 0) \).

- **Minimum value \( c = 20 \)**
- **Occurs at the endpoints:**
  - **Smaller \( x \)-value**: \( (6, 4) \)
  - **Larger \( x \)-value**: \( (10, 0) \)

The feasible region is bounded, so there is an optimal solution.

Would you like further explanation on any of these steps?

---

1. How can we determine if a feasible region is bounded or unbounded in an LP problem?
2. Why do we check each vertex in the feasible region when minimizing or maximizing in LP problems?
3. How does dividing each constraint by a common factor help in simplifying inequalities?
4. What methods exist beyond substitution to find intersections in linear programming?
5. How does an objective function differ in sensitivity to changes at each vertex?

**Tip:** In LP problems, always simplify constraints first; this helps identify intersections quickly and accurately.

Solve the LP problem. If no optimal solution exists, indicate whether the feasible region is empty or the objective function is unbounded. (Enter EMPTY if the region is empty. Enter UNBOUNDED if the function is unbounded.)
Minimize c = 2x + 4y subject to
0.3x + 0.3y ≥ 3
3x + 6y ≥ 42
x ≥ 0, y ≥ 0.
Minimum value c = 
 occurs at the endpoints 
(x, y) = 

 


 (smaller x-value) and 
(x, y) = 

 


 (larger x-value) and 
---Select--- .

Step-by-step solution to a linear programming problem involving the minimization of c = 2x + 4y under given constraints. Includes methods for identifying feasible regions, verifying boundedness, and calculating the minimum value at specific vertices. Suitable for college-level students studying optimization.

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