A diet for a sick person must contain at least 4000 units of vitamins, 50 units of minerals and 1400 units of calories. Two foods A and B are available at a cost of Rs. 5 and Rs. 4 per unit respectively. One unit of food A contains 200 units of vitamins, 1 unit of minerals and 40 units of calories whereas one unit of food B contains 100 units of vitamins, 2 units of minerals and 40 units of calories. Find what combination of the food A and B should be used to have least cost but it must satisfy the requirements of the sick person.

Answer

To solve the problem of finding the optimal combination of food A and food B for a sick person's diet while minimizing costs, we will follow these steps:

### Step 1: Define Variables
Let:
- \( x \) = number of units of food A
- \( y \) = number of units of food B

### Step 2: Formulate the Objective Function
The cost of food A is Rs. 5 per unit, and the cost of food B is Rs. 4 per unit. Therefore, the objective function to minimize the cost \( Z \) is:
\[ Z = 5x + 4y \]

### Step 3: Set Up the Constraints
Based on the nutritional requirements given in the problem:
1. **Vitamins**: Each unit of food A provides 200 units of vitamins, and each unit of food B provides 100 units. The requirement is at least 4000 units of vitamins:
\[ 200x + 100y \geq 4000 \]
Simplifying this gives:
\[ 2x + y \geq 40 \quad \text{(Constraint 1)} \]

2. **Minerals**: Each unit of food A provides 1 unit of minerals, and each unit of food B provides 2 units. The requirement is at least 50 units of minerals:
\[ x + 2y \geq 50 \quad \text{(Constraint 2)} \]

3. **Calories**: Each unit of food A provides 40 units of calories, and each unit of food B provides 40 units. The requirement is at least 1400 units of calories:
\[ 40x + 40y \geq 1400 \]
Simplifying this gives:
\[ x + y \geq 35 \quad \text{(Constraint 3)} \]

### Step 4: Graph the Constraints
We will graph the inequalities defined by the constraints to find the feasible region.

1. For \( 2x + y = 40 \):
- When \( x = 0 \), \( y = 40 \) (Point A)
- When \( y = 0 \), \( x = 20 \) (Point B)

2. For \( x + 2y = 50 \):
- When \( x = 0 \), \( y = 25 \) (Point C)
- When \( y = 0 \), \( x = 50 \) (Point D)

3. For \( x + y = 35 \):
- When \( x = 0 \), \( y = 35 \) (Point E)
- When \( y = 0 \), \( x = 35 \) (Point F)

### Step 5: Identify the Feasible Region
The feasible region is the area where all constraints overlap, and it is bounded by the lines we plotted. We will only consider the first quadrant since \( x \) and \( y \) must be non-negative.

### Step 6: Find the Corner Points of the Feasible Region
We need to find the intersection points of the lines to determine the corner points of the feasible region.

1. **Intersection of \( 2x + y = 40 \) and \( x + y = 35 \)**:
- Solve the equations:
\[ y = 35 - x \]
Substitute into \( 2x + (35 - x) = 40 \):
\[ 2x + 35 - x = 40 \implies x = 5, \quad y = 30 \quad \text{(Point B: (5, 30))} \]

2. **Intersection of \( x + 2y = 50 \) and \( x + y = 35 \)**:
- Solve the equations:
\[ y = 35 - x \]
Substitute into \( x + 2(35 - x) = 50 \):
\[ x + 70 - 2x = 50 \implies x = 20, \quad y = 15 \quad \text{(Point C: (20, 15))} \]

3. **Intersection of \( 2x + y = 40 \) and \( x + 2y = 50 \)**:
- Solve the equations:
\[ y = 40 - 2x \]
Substitute into \( x + 2(40 - 2x) = 50 \):
\[ x + 80 - 4x = 50 \implies 3x = 30 \implies x = 10, \quad y = 20 \quad \text{(Point D: (10, 20))} \]

### Step 7: Evaluate the Objective Function at Each Corner Point
Now we will evaluate the cost function \( Z = 5x + 4y \) at each corner point:

1. **At Point A (0, 40)**:
\[ Z = 5(0) + 4(40) = 160 \]

2. **At Point B (5, 30)**:
\[ Z = 5(5) + 4(30) = 25 + 120 = 145 \]

3. **At Point C (20, 15)**:
\[ Z = 5(20) + 4(15) = 100 + 60 = 160 \]

4. **At Point D (10, 20)**:
\[ Z = 5(10) + 4(20) = 50 + 80 = 130 \]

### Step 8: Determine the Minimum Cost
The minimum cost occurs at Point D (10, 20) with a cost of Rs. 130.

### Final Answer
To minimize the cost while meeting the dietary requirements, the sick person should consume:
- **10 units of food A**
- **20 units of food B**
- **Minimum cost is Rs. 130**

---