To solve the problem step-by-step, we will set up the equations based on the information provided and then find the maximum profit.
### Step 1: Define the variables
Let:
- \( x \) = number of shares of type \( s_1 \)
- \( y \) = number of shares of type \( s_2 \)
### Step 2: Set up the constraints
1. The total number of shares purchased cannot exceed 10:
\[
x + y \leq 10 \quad \text{(Constraint 1)}
\]
2. The total cost of the shares cannot exceed Rs. 1500:
\[
180x + 120y \leq 1500 \quad \text{(Constraint 2)}
\]
### Step 3: Simplify the cost constraint
To simplify Constraint 2, divide the entire inequality by 60:
\[
3x + 2y \leq 25 \quad \text{(Constraint 2 simplified)}
\]
### Step 4: Define the objective function
The profit from the shares is given by:
\[
Z = 11x + 8y \quad \text{(Objective function)}
\]
We need to maximize \( Z \).
### Step 5: Identify the feasible region
We will graph the constraints to find the feasible region.
- For Constraint 1: \( x + y = 10 \)
- If \( x = 0 \), then \( y = 10 \) (point A: (0, 10))
- If \( y = 0 \), then \( x = 10 \) (point C: (10, 0))
- For Constraint 2: \( 3x + 2y = 25 \)
- If \( x = 0 \), then \( y = 12.5 \) (not feasible since \( y \) must be ≤ 10)
- If \( y = 0 \), then \( x = \frac{25}{3} \approx 8.33 \) (point B: (8.33, 0))
- If \( y = 10 \), then \( 3x + 20 = 25 \) gives \( x = \frac{5}{3} \approx 1.67 \) (point D: (1.67, 10))
### Step 6: Determine the intersection point of the constraints
To find the intersection of \( x + y = 10 \) and \( 3x + 2y = 25 \):
1. From \( x + y = 10 \), express \( y \):
\[
y = 10 - x
\]
2. Substitute into \( 3x + 2y = 25 \):
\[
3x + 2(10 - x) = 25
\]
\[
3x + 20 - 2x = 25
\]
\[
x = 5
\]
3. Substitute \( x = 5 \) back to find \( y \):
\[
y = 10 - 5 = 5
\]
Thus, the intersection point is (5, 5).
### Step 7: Evaluate the objective function at the vertices of the feasible region
Now, we evaluate \( Z \) at the vertices:
1. Point A (0, 10):
\[
Z = 11(0) + 8(10) = 80
\]
2. Point B (5, 5):
\[
Z = 11(5) + 8(5) = 55 + 40 = 95
\]
3. Point C (8.33, 0):
\[
Z = 11(8.33) + 8(0) \approx 91.63
\]
4. Point D (1.67, 10):
\[
Z = 11(1.67) + 8(10) \approx 13.37 + 80 = 93.37
\]
### Step 8: Determine the maximum profit
The maximum profit occurs at point B (5, 5) where:
\[
Z = 95
\]
### Conclusion
The man should purchase:
- **5 shares of type \( s_1 \)**
- **5 shares of type \( s_2 \)**
The maximum profit will be **Rs. 95**.
