the members of the cat appreciation club on campus are selling shirts (x) and hats (y) and they want to maximize their profit. objective function: p=2x+y. Constraints: x+2y ≤ 5, x-y ≤2, x≥0, and y≥0. First, graph each system of constraints to find the feasible region. remember to shade in the overlapping areas. then name the vertices of te feasible region. then solve to find the values of x and y that maximize the profit objective function, WHAT IS THE MAXIMUM PROFIT?

To solve this problem, we need to graph the system of inequalities, find the feasible region, identify its vertices, and then use these vertices to determine the values of \(x\) and \(y\) that maximize the profit function \(P = 2x + y\).

### Step 1: Graph the System of Constraints

The constraints given are:

1. \(x + 2y \leq 5\)
2. \(x - y \leq 2\)
3. \(x \geq 0\)
4. \(y \geq 0\)

Let's graph these inequalities:

#### 1. Graph \(x + 2y \leq 5\):
- The boundary line is \(x + 2y = 5\).
- To find the intercepts:
  - Set \(x = 0\): \(2y = 5 \implies y = \frac{5}{2}\).
  - Set \(y = 0\): \(x = 5\).

This line passes through \((0, \frac{5}{2})\) and \((5, 0)\).

#### 2. Graph \(x - y \leq 2\):
- The boundary line is \(x - y = 2\).
- To find the intercepts:
  - Set \(x = 0\): \(-y = 2 \implies y = -2\) (but since \(y \geq 0\), this point is not in the feasible region).
  - Set \(y = 0\): \(x = 2\).

This line passes through \((2, 0)\) and would intersect the y-axis at \((0, -2)\).

#### 3. Graph \(x \geq 0\) and \(y \geq 0\):
- These are the non-negative constraints, so we only consider the first quadrant.

### Step 2: Find the Feasible Region
The feasible region is the overlapping area that satisfies all constraints.

### Step 3: Identify the Vertices of the Feasible Region
To identify the vertices of the feasible region, find the intersection points of the boundary lines within the first quadrant:

1. Intersection of \(x + 2y = 5\) and \(x - y = 2\):
   - Solve the system:
     \[
     x + 2y = 5 \quad (1)\\
     x - y = 2 \quad (2)
     \]
   - From (2): \(x = y + 2\).
   - Substitute in (1): \((y + 2) + 2y = 5\) \(\Rightarrow 3y + 2 = 5\) \(\Rightarrow 3y = 3\) \(\Rightarrow y = 1\).
   - Then, \(x = 1 + 2 = 3\).
   - So, intersection is \((3, 1)\).

2. Intersection of \(x + 2y = 5\) and \(x = 0\):
   - Set \(x = 0\) in \(x + 2y = 5\): \(2y = 5 \implies y = \frac{5}{2}\).
   - So, intersection is \((0, \frac{5}{2})\).

3. Intersection of \(x - y = 2\) and \(y = 0\):
   - Set \(y = 0\) in \(x - y = 2\): \(x = 2\).
   - So, intersection is \((2, 0)\).

4. Intersection of \(x = 0\) and \(y = 0\) is the origin \((0, 0)\).

### Step 4: Evaluate the Profit Function at Each Vertex
Now we evaluate the objective function \(P = 2x + y\) at each vertex:

1. At \((0, 0)\): \(P = 2(0) + 0 = 0\).
2. At \((2, 0)\): \(P = 2(2) + 0 = 4\).
3. At \((0, \frac{5}{2})\): \(P = 2(0) + \frac{5}{2} = 2.5\).
4. At \((3, 1)\): \(P = 2(3) + 1 = 7\).

### Step 5: Find the Maximum Profit
The maximum profit is at the point \((3, 1)\) where \(P = 7\).

### Conclusion
The values of \(x = 3\) shirts and \(y = 1\) hat maximize the profit, and the **maximum profit is \(7\) units**.

Would you like a detailed graph or have any other questions? 

### Additional Questions:
1. What happens if the price per shirt or hat changes?
2. How would the feasible region change if an additional constraint were added?
3. What if the profit function was different, such as \(P = x + 3y\)?
4. Can you interpret the significance of each vertex in a business context?
5. How does altering constraints impact the feasible region and profit?

**Tip:** When solving optimization problems, always verify the constraints and objective function are set correctly to ensure accurate results.

Explore how to maximize profit by selling shirts and hats using linear programming. Graph the constraints, identify the feasible region vertices, and solve the profit objective function. Learn the steps and calculations involved in optimizing profit in this mathematical problem.

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