To solve the problem, we need to find the combination of foods X and Y that minimizes the cost while satisfying the dietary requirements. Let's break it down step by step.
### Step 1: Define the Variables
Let:
- \( x \) = number of units of food X
- \( y \) = number of units of food Y
### Step 2: Define the Objective Function
The cost for food X is ₹4 per unit and for food Y is ₹3 per unit. Therefore, the total cost \( z \) can be defined as:
\[
z = 4x + 3y
\]
### Step 3: Define the Constraints
We have three dietary requirements that must be satisfied:
1. Vitamins: Each unit of food X contains 200 units of vitamins, and each unit of food Y contains 100 units. The total vitamins must be at least 4000 units:
\[
200x + 100y \geq 4000
\]
Simplifying this, we divide by 100:
\[
2x + y \geq 40 \quad \text{(Constraint 1)}
\]
2. Minerals: Each unit of food X contains 1 unit of minerals, and each unit of food Y contains 2 units. The total minerals must be at least 50 units:
\[
x + 2y \geq 50 \quad \text{(Constraint 2)}
\]
3. Calories: Each unit of food X contains 40 calories, and each unit of food Y contains 40 calories. The total calories must be at least 1400:
\[
40x + 40y \geq 1400
\]
Simplifying this, we divide by 40:
\[
x + y \geq 35 \quad \text{(Constraint 3)}
\]
### Step 4: Rewrite the Constraints
Now we have the following constraints:
1. \( 2x + y \geq 40 \)
2. \( x + 2y \geq 50 \)
3. \( x + y \geq 35 \)
### Step 5: Find the Corner Points
To find the feasible region, we will solve the equations of the constraints to find the corner points.
1. From \( 2x + y = 40 \):
- If \( x = 0 \), then \( y = 40 \) (Point A: (0, 40))
- If \( y = 0 \), then \( x = 20 \) (Point B: (20, 0))
2. From \( x + 2y = 50 \):
- If \( x = 0 \), then \( y = 25 \) (Point C: (0, 25))
- If \( y = 0 \), then \( x = 50 \) (Point D: (50, 0))
3. From \( x + y = 35 \):
- If \( x = 0 \), then \( y = 35 \) (Point E: (0, 35))
- If \( y = 0 \), then \( x = 35 \) (Point F: (35, 0))
### Step 6: Find Intersection Points
Now we need to find the intersection points of the constraints:
1. Solve \( 2x + y = 40 \) and \( x + 2y = 50 \):
- From \( 2x + y = 40 \), we can express \( y = 40 - 2x \).
- Substitute into \( x + 2(40 - 2x) = 50 \):
\[
x + 80 - 4x = 50 \implies -3x = -30 \implies x = 10
\]
- Substitute \( x = 10 \) back into \( y = 40 - 2(10) = 20 \).
- Intersection point: (10, 20)
2. Solve \( x + 2y = 50 \) and \( x + y = 35 \):
- From \( x + y = 35 \), we can express \( y = 35 - x \).
- Substitute into \( x + 2(35 - x) = 50 \):
\[
x + 70 - 2x = 50 \implies -x = -20 \implies x = 20
\]
- Substitute \( x = 20 \) back into \( y = 35 - 20 = 15 \).
- Intersection point: (20, 15)
3. Solve \( 2x + y = 40 \) and \( x + y = 35 \):
- From \( x + y = 35 \), we can express \( y = 35 - x \).
- Substitute into \( 2x + (35 - x) = 40 \):
\[
2x + 35 - x = 40 \implies x = 5
\]
- Substitute \( x = 5 \) back into \( y = 35 - 5 = 30 \).
- Intersection point: (5, 30)
### Step 7: Evaluate the Objective Function at Each Corner Point
Now we evaluate \( z = 4x + 3y \) at each corner point:
1. At (0, 40): \( z = 4(0) + 3(40) = 120 \)
2. At (5, 30): \( z = 4(5) + 3(30) = 20 + 90 = 110 \)
3. At (20, 15): \( z = 4(20) + 3(15) = 80 + 45 = 125 \)
4. At (50, 0): \( z = 4(50) + 3(0) = 200 \)
### Step 8: Determine the Minimum Cost
The minimum cost occurs at the point (5, 30) with a cost of ₹110.
### Final Answer
The combination of foods X and Y that should be used to minimize cost while satisfying the dietary requirements is:
- \( x = 5 \) units of food X
- \( y = 30 \) units of food Y
