To solve the problem, we need to formulate a Linear Programming Problem (LPP) based on the given information about the food items and their vitamin content. Here’s how we can approach it step by step:
### Step 1: Define Variables
Let:
- \( x \) = amount (in kg) of food \( F_1 \)
- \( y \) = amount (in kg) of food \( F_2 \)
### Step 2: Set Up the Objective Function
The objective is to minimize the cost of the mixture. The cost per kg of food \( F_1 \) is ₹6 and for food \( F_2 \) is ₹10. Therefore, the objective function can be expressed as:
\[
\text{Minimize } Z = 6x + 10y
\]
### Step 3: Set Up the Constraints
From the problem, we know the vitamin content in each food item:
- **Vitamin A**: Food \( F_1 \) contributes 1 unit and Food \( F_2 \) contributes 2 units. We need at least 10 units:
\[
1x + 2y \geq 10
\]
- **Vitamin B**: Food \( F_1 \) contributes 2 units and Food \( F_2 \) contributes 2 units. We need at least 12 units:
\[
2x + 2y \geq 12
\]
- **Vitamin C**: Food \( F_1 \) contributes 3 units and Food \( F_2 \) contributes 1 unit. We need at least 8 units:
\[
3x + 1y \geq 8
\]
Additionally, we have non-negativity constraints:
\[
x \geq 0, \quad y \geq 0
\]
### Step 4: Formulate the LPP
Putting it all together, we have the following Linear Programming Problem:
\[
\text{Minimize } Z = 6x + 10y
\]
Subject to:
\[
\begin{align*}
1x + 2y & \geq 10 \quad \text{(1)} \\
2x + 2y & \geq 12 \quad \text{(2)} \\
3x + 1y & \geq 8 \quad \text{(3)} \\
x & \geq 0 \\
y & \geq 0
\end{align*}
\]
### Step 5: Solve the Constraints
To find the feasible region, we will convert the inequalities into equations and find the intersection points.
1. From \( 1x + 2y = 10 \):
- If \( x = 0 \), then \( y = 5 \) (Point A: (0, 5))
- If \( y = 0 \), then \( x = 10 \) (Point B: (10, 0))
2. From \( 2x + 2y = 12 \):
- If \( x = 0 \), then \( y = 6 \) (Point C: (0, 6))
- If \( y = 0 \), then \( x = 6 \) (Point D: (6, 0))
3. From \( 3x + 1y = 8 \):
- If \( x = 0 \), then \( y = 8 \) (Point E: (0, 8))
- If \( y = 0 \), then \( x = \frac{8}{3} \approx 2.67 \) (Point F: (2.67, 0))
### Step 6: Find Intersection Points
To find the intersection points of the lines, we solve the equations pairwise.
1. **Intersection of (1) and (2)**:
\[
\begin{align*}
x + 2y &= 10 \\
2x + 2y &= 12
\end{align*}
\]
Subtracting the first from the second gives:
\[
x = 2 \\
\text{Substituting } x = 2 \text{ in } x + 2y = 10 \Rightarrow 2 + 2y = 10 \Rightarrow 2y = 8 \Rightarrow y = 4
\]
Intersection point: (2, 4)
2. **Intersection of (1) and (3)**:
\[
\begin{align*}
x + 2y &= 10 \\
3x + y &= 8
\end{align*}
\]
Solving these gives:
\[
y = 10 - x \quad \Rightarrow \quad 3x + (10 - x) = 8 \Rightarrow 2x = -2 \Rightarrow x = 1, y = 9
\]
Intersection point: (1, 9)
3. **Intersection of (2) and (3)**:
\[
\begin{align*}
2x + 2y &= 12 \\
3x + y &= 8
\end{align*}
\]
Solving these gives:
\[
y = 8 - 3x \quad \Rightarrow \quad 2x + 2(8 - 3x) = 12 \Rightarrow -4x + 16 = 12 \Rightarrow x = 1
\]
Intersection point: (1, 5)
### Step 7: Evaluate the Objective Function at Each Vertex
Now we evaluate \( Z = 6x + 10y \) at each of the vertices found:
1. At (0, 5): \( Z = 6(0) + 10(5) = 50 \)
2. At (2, 4): \( Z = 6(2) + 10(4) = 12 + 40 = 52 \)
3. At (1, 5): \( Z = 6(1) + 10(5) = 6 + 50 = 56 \)
### Step 8: Conclusion
The minimum cost occurs at the point (0, 5) with a cost of ₹50.
