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 + 2y subject to
x + 3y	≥	22
4x + y	≥	22
x ≥ 0, y ≥ 0.
c	=	
 

(x, y)	
= 

  

 



To solve this linear programming (LP) problem, let's break it down systematically.

### Step 1: Analyze the constraints
The problem is:
- Minimize \( c = x + 2y \)
- Subject to:
  1. \( x + 3y \geq 22 \)
  2. \( 4x + y \geq 22 \)
  3. \( x \geq 0 \)
  4. \( y \geq 0 \)

These constraints define a feasible region in the first quadrant of the coordinate plane where both \( x \) and \( y \) are non-negative.

### Step 2: Rewrite inequalities as equalities to find boundary lines
1. \( x + 3y = 22 \)
   - Solving for \( y \): \( y = \frac{22 - x}{3} \)
2. \( 4x + y = 22 \)
   - Solving for \( y \): \( y = 22 - 4x \)

### Step 3: Find the intersection points of the boundary lines
- **Intersection of** \( x + 3y = 22 \) **and** \( 4x + y = 22 \):
  \[
  x + 3y = 22
  \]
  \[
  4x + y = 22
  \]
  
  Solving this system by substitution or elimination yields:
  - Multiply the first equation by 4: \( 4x + 12y = 88 \)
  - Subtract the second equation: \( 11y = 66 \Rightarrow y = 6 \)
  - Substitute \( y = 6 \) into \( x + 3y = 22 \): \( x + 18 = 22 \Rightarrow x = 4 \)

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

### Step 4: Check vertices
The vertices of the feasible region where constraints are met will be the points:
1. The intersection point \( (4, 6) \).
2. Where the lines intersect the axes (subject to \( x, y \geq 0 \)).

Evaluating each:
- **On \( x + 3y = 22 \) with \( x = 0 \):** \( y = \frac{22}{3} \approx 7.33 \)
- **On \( 4x + y = 22 \) with \( y = 0 \):** \( x = \frac{22}{4} = 5.5 \)

So, the feasible region vertices are approximately:
- \( (0, 7.33) \)
- \( (5.5, 0) \)
- \( (4, 6) \)

### Step 5: Evaluate the objective function at vertices
Calculate \( c = x + 2y \) at each vertex:
1. At \( (0, 7.33) \): \( c = 0 + 2(7.33) = 14.66 \)
2. At \( (5.5, 0) \): \( c = 5.5 + 2(0) = 5.5 \)
3. At \( (4, 6) \): \( c = 4 + 2(6) = 16 \)

### Step 6: Identify the minimum
The minimum value of \( c \) occurs at \( (5.5, 0) \) with \( c = 5.5 \).

### Answer
\[
c = 5.5
\]
\[
(x, y) = (5.5, 0)
\]

Would you like a detailed graph or further clarification?

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**Relative Questions:**
1. How do we graphically represent feasible regions in linear programming?
2. What does it mean for an LP problem to be unbounded?
3. How can we check if the objective function has a unique minimum or multiple solutions?
4. What role do intersection points play in identifying feasible solutions in LP?
5. How does non-negativity restriction affect the feasible region?

**Tip:** In LP problems, always plot constraints to visualize the feasible region easily.

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 + 2y subject to x + 3y ≥ 22, 4x + y ≥ 22, x ≥ 0, y ≥ 0.

Solve this linear programming problem by minimizing the objective function c = x + 2y with constraints x + 3y ≥ 22, 4x + y ≥ 22, x ≥ 0, y ≥ 0. This solution covers key LP concepts such as feasible region, boundary lines, and vertices evaluation. Ideal for college-level or advanced high school students.

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