CONSIDER THE FOLLOWING LINEAR PROGRAM:
Minimize C = 75P + 32Q
Subject To:
                9P + 5Q >= 45        (1)
                           Q <= 6         (2)
                           P <=3           (3)
                P>=0, Q>=0 
If you place P in the horizontal axis, which of the following is NOT a vertex of the set of feasible solutions?

We are tasked with determining which point is **not** a vertex of the set of feasible solutions for the given linear program:

### Objective:
Minimize:  
\[ C = 75P + 32Q \]

### Subject to:
1. \( 9P + 5Q \geq 45 \)
2. \( Q \leq 6 \)
3. \( P \leq 3 \)
4. \( P \geq 0, Q \geq 0 \)

### Step 1: Graph the inequalities and determine the feasible region.
- **Inequality (1):**  
  \( 9P + 5Q \geq 45 \)  
  This can be rearranged into equality form for graphing:
  \[ Q = \frac{45 - 9P}{5} \]  
  This represents a line, and the inequality defines the region **above** the line.
  
- **Inequality (2):**  
  \( Q \leq 6 \)  
  This defines a horizontal line at \( Q = 6 \), and the region is **below** or on this line.
  
- **Inequality (3):**  
  \( P \leq 3 \)  
  This defines a vertical line at \( P = 3 \), and the region is to the **left** of this line.

- **Non-negativity constraints:**  
  \( P \geq 0 \) and \( Q \geq 0 \), confining the region to the first quadrant.

### Step 2: Find the vertices by solving system of equations.
We solve for the intersection points of the constraints.

#### 1. Intersection of \( 9P + 5Q = 45 \) and \( Q = 6 \):
Substitute \( Q = 6 \) into the first constraint:
\[
9P + 5(6) = 45
\]
\[
9P + 30 = 45 \implies 9P = 15 \implies P = \frac{15}{9} = \frac{5}{3}
\]
Thus, the intersection point is \( \left( \frac{5}{3}, 6 \right) \).

#### 2. Intersection of \( 9P + 5Q = 45 \) and \( P = 3 \):
Substitute \( P = 3 \) into the first constraint:
\[
9(3) + 5Q = 45
\]
\[
27 + 5Q = 45 \implies 5Q = 18 \implies Q = \frac{18}{5}
\]
Thus, the intersection point is \( \left( 3, \frac{18}{5} \right) \).

#### 3. Intersection of \( P = 3 \) and \( Q = 6 \):
The point is \( (3, 6) \).

#### 4. Intersection of \( P = 0 \) and \( 9P + 5Q = 45 \):
Substitute \( P = 0 \) into the first constraint:
\[
9(0) + 5Q = 45 \implies 5Q = 45 \implies Q = 9
\]
Thus, the intersection point is \( (0, 9) \).

#### 5. Intersection of \( P = 0 \) and \( Q = 6 \):
The point is \( (0, 6) \).

### Step 3: List of potential vertices.
The vertices of the feasible region are:
- \( \left( \frac{5}{3}, 6 \right) \)
- \( (3, \frac{18}{5}) \)
- \( (3, 6) \)
- \( (0, 6) \)

### Step 4: Identify the point that is **not** a vertex.
From the list of vertices, it is clear that \( (0, 9) \) is **not** a vertex of the feasible region, as it does not satisfy the constraint \( Q \leq 6 \).

Thus, the point **\( (0, 9) \)** is **not** a vertex of the feasible solution set.

---

Would you like further clarification on any step? Here are some follow-up questions:

1. How do you solve systems of linear inequalities graphically?
2. Can you explain how vertices are critical in linear programming?
3. How does changing constraints affect the feasible region?
4. Why is the non-negativity condition important in linear programming?
5. How do objective functions affect which vertex is optimal?

**Tip:** When graphing inequalities, always check which side of the boundary line satisfies the inequality.

CONSIDER THE FOLLOWING LINEAR PROGRAM: Minimize C = 75P + 32Q Subject To: 9P + 5Q >= 45 (1) Q <= 6 (2) P <= 3 (3) P >= 0, Q >= 0 If you place P in the horizontal axis, which of the following is NOT a vertex of the set of feasible solutions?

Explore a linear programming problem with constraints 9P + 5Q >= 45, Q <= 6, and P <= 3, focusing on graphing inequalities, determining the feasible region, and finding which point is not a vertex. Ideal for advanced students interested in optimization techniques.

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