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* = *x* + 2*y* subject to

*x* + 9*y*

≥

26

3*x* + *y*

≥

26

*x* ≥ 0, *y* ≥ 0.

*c*= (*x*, *y*)

= 

We are asked to solve the following linear programming (LP) problem:

**Objective:**
Minimize \( c = x + 2y \)

**Subject to the constraints:**
1. \( x + 9y \geq 26 \)
2. \( 3x + y \geq 26 \)
3. \( x \geq 0 \)
4. \( y \geq 0 \)

### Step 1: Analyze the constraints graphically
Each inequality represents a half-plane in the \( xy \)-plane. We need to find the region where all the constraints hold simultaneously.

#### 1. \( x + 9y \geq 26 \)
This can be rewritten as:
\[ y \geq \frac{26 - x}{9} \]
This is a line with slope \( -\frac{1}{9} \) and y-intercept \( \frac{26}{9} \approx 2.89 \).

#### 2. \( 3x + y \geq 26 \)
This can be rewritten as:
\[ y \geq 26 - 3x \]
This is a line with slope \( -3 \) and y-intercept 26.

#### 3. \( x \geq 0 \)
This represents the region to the right of the \( y \)-axis.

#### 4. \( y \geq 0 \)
This represents the region above the \( x \)-axis.

### Step 2: Find the intersection of constraints
The feasible region is where all the constraints overlap.

#### Intersection of \( x + 9y = 26 \) and \( 3x + y = 26 \):
We solve these two equations to find the point of intersection.

1. From the first equation \( x + 9y = 26 \), solve for \( x \):
   \[ x = 26 - 9y \]

2. Substitute this into the second equation \( 3x + y = 26 \):
   \[ 3(26 - 9y) + y = 26 \]
   \[ 78 - 27y + y = 26 \]
   \[ -26y = -52 \]
   \[ y = 2 \]

3. Substitute \( y = 2 \) back into \( x = 26 - 9y \):
   \[ x = 26 - 9(2) = 26 - 18 = 8 \]

So, the point of intersection is \( (8, 2) \).

### Step 3: Check the boundaries
We now check the boundaries of the feasible region by considering the constraints at extreme points.

- At \( y = 0 \):
  - From \( x + 9y = 26 \), we get \( x = 26 \), giving the point \( (26, 0) \).
  - From \( 3x + y = 26 \), we get \( x = \frac{26}{3} \approx 8.67 \), giving the point \( (8.67, 0) \).

The feasible region lies between these extreme points, including the point of intersection \( (8, 2) \).

### Step 4: Evaluate the objective function at corner points
Now, evaluate \( c = x + 2y \) at the corner points of the feasible region.

1. At \( (26, 0) \):
   \[ c = 26 + 2(0) = 26 \]

2. At \( (8, 2) \):
   \[ c = 8 + 2(2) = 8 + 4 = 12 \]

3. At \( (8.67, 0) \):
   \[ c = 8.67 + 2(0) = 8.67 \]

### Step 5: Conclusion
The minimum value of \( c \) occurs at the point \( (8, 2) \) with \( c = 12 \).

Thus, the optimal solution is:
\[
\boxed{12}
\]

Would you like more details or further explanations?

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Here are 5 relative questions to deepen your understanding:

1. How do we determine whether the objective function is unbounded in an LP problem?
2. What is the geometric significance of the feasible region in linear programming?
3. How does changing the objective function affect the location of the optimal solution?
4. Can an LP problem have multiple optimal solutions? Under what conditions?
5. What methods can be used to solve LP problems besides the graphical approach?

**Tip:** In LP, the feasible region is always a convex set, which means the optimal solution is found at a vertex (corner point) if it exists.

Solve the LP problem. If no optimal solution exists, indicate whether the feasible region is empty or the objective function is unbounded. Minimize c = x + 2y subject to: x + 9y ≥ 26, 3x + y ≥ 26, x ≥ 0, y ≥ 0.

Solve the linear programming problem by minimizing c = x + 2y with constraints x + 9y ≥ 26 and 3x + y ≥ 26. Learn how to find the optimal solution using the graphical method and analyze the feasible region. Ideal for advanced high school or college students.

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