A hospital dietician wishes to find the cheapest combination of two foods, A and B that contains at least 0.5 milligram of thiamin and at least 600 calories. Each unit of A contains 0.12 milligram of thiamin and 100 calories, while each unit of B contains 0.10 milligram of thiamin and 150 calories. If each food costs Rs. 10 per unit, how many units of each should be combined at a minimum cost?

Answer

To solve the problem of finding the cheapest combination of foods A and B that meets the dietary requirements, we can follow these steps:

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

### Step 2: Set Up the Inequalities
According to the problem, we need to satisfy the following conditions:
1. Thiamin requirement:
\[ 0.12x + 0.10y \geq 0.5 \]
2. Caloric requirement:
\[ 100x + 150y \geq 600 \]

### Step 3: Set Up the Objective Function
We want to minimize the cost, which is given by:
\[ \text{Cost} = 10x + 10y \]
This can be simplified to:
\[ \text{Cost} = 10(x + y) \]

### Step 4: Solve the Inequalities
We will solve the inequalities to find feasible values for \( x \) and \( y \).

**From the Thiamin inequality:**
\[ 0.12x + 0.10y \geq 0.5 \]
Multiply through by 100 to eliminate decimals:
\[ 12x + 10y \geq 50 \]

**From the Caloric inequality:**
\[ 100x + 150y \geq 600 \]
Divide through by 50:
\[ 2x + 3y \geq 12 \]

### Step 5: Graph the Inequalities
To find the feasible region, graph the lines:
1. \( 12x + 10y = 50 \)
2. \( 2x + 3y = 12 \)

**Finding intercepts for \( 12x + 10y = 50 \):**
- If \( x = 0 \): \( 10y = 50 \) → \( y = 5 \) (y-intercept)
- If \( y = 0 \): \( 12x = 50 \) → \( x \approx 4.17 \) (x-intercept)

**Finding intercepts for \( 2x + 3y = 12 \):**
- If \( x = 0 \): \( 3y = 12 \) → \( y = 4 \) (y-intercept)
- If \( y = 0 \): \( 2x = 12 \) → \( x = 6 \) (x-intercept)

### Step 6: Identify the Feasible Region
The feasible region is where the inequalities overlap.

### Step 7: Test Corner Points
Evaluate the objective function \( 10(x + y) \) at the corner points of the feasible region:
1. \( (0, 5) \): Cost = \( 10(0 + 5) = 50 \)
2. \( (4.17, 0) \): Cost = \( 10(4.17 + 0) \approx 41.7 \)
3. \( (0, 4) \): Cost = \( 10(0 + 4) = 40 \)
4. \( (2, 3) \): Cost = \( 10(2 + 3) = 50 \)

### Step 8: Conclusion
The minimum cost occurs at the point \( (2, 3) \), which means:
- 2 units of food A
- 3 units of food B

### Final Answer
The cheapest combination is 2 units of food A and 3 units of food B, costing Rs. 50.

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