A woman wishes to mix two types of foods in such a way that the vitamin contents of the mixture contain at least 8 units of vitamin A and at least 10 units of vitamin C. Food `I` contains 2 units/kg of vitamin A and 1 unit /kg of vitamin C. Food II contains 1 unit/kg of vitamin A and 2 unit/kg of vitamin C. The cost of food I is Rs. 50/kg and of food II is Rs. 70/kg. Find the minimum cost of such a mixture.

Explanation

To solve the problem of mixing two types of foods to meet vitamin requirements at minimum cost, we will follow these steps:

### Step 1: Define Variables
Let:
- \( x \) = kg of Food I
- \( y \) = kg of Food II

### Step 2: Set Up the Constraints
From the problem, we know:
- Food I contains 2 units of Vitamin A and 1 unit of Vitamin C.
- Food II contains 1 unit of Vitamin A and 2 units of Vitamin C.

The requirements for vitamins are:
1. At least 8 units of Vitamin A:
\[ 2x + y \geq 8 \]
2. At least 10 units of Vitamin C:
\[ x + 2y \geq 10 \]

Additionally, we have non-negativity constraints:
\[ x \geq 0, \quad y \geq 0 \]

### Step 3: Set Up the Objective Function
The cost of the foods is given as:
- Cost of Food I = Rs. 50/kg
- Cost of Food II = Rs. 70/kg

Thus, the objective function to minimize the total cost \( Z \) is:
\[ Z = 50x + 70y \]

### Step 4: Graph the Constraints
To graph the constraints, we can convert the inequalities into equations:

1. For \( 2x + y = 8 \):
- If \( x = 0 \), then \( y = 8 \) (point (0, 8))
- If \( y = 0 \), then \( x = 4 \) (point (4, 0))

2. For \( x + 2y = 10 \):
- If \( x = 0 \), then \( y = 5 \) (point (0, 5))
- If \( y = 0 \), then \( x = 10 \) (point (10, 0))

### Step 5: Identify the Feasible Region
Plot the lines on a graph and shade the feasible region that satisfies both inequalities. The feasible region will be bounded by the lines and the axes.

### Step 6: Find Corner Points
The corner points of the feasible region can be found by solving the equations of the lines:
1. Intersection of \( 2x + y = 8 \) and \( x + 2y = 10 \):
- From \( 2x + y = 8 \), express \( y \):
\[ y = 8 - 2x \]
- Substitute into \( x + 2y = 10 \):
\[ x + 2(8 - 2x) = 10 \implies x + 16 - 4x = 10 \implies -3x = -6 \implies x = 2 \]
\[ y = 8 - 2(2) = 4 \]
- So, one corner point is \( (2, 4) \).

2. The other corner points are \( (0, 8) \) and \( (4, 0) \).

### Step 7: Evaluate the Objective Function at Corner Points
Now, we will evaluate the cost function \( Z = 50x + 70y \) at each corner point:
1. At \( (0, 8) \):
\[ Z = 50(0) + 70(8) = 560 \]
2. At \( (2, 4) \):
\[ Z = 50(2) + 70(4) = 100 + 280 = 380 \]
3. At \( (4, 0) \):
\[ Z = 50(4) + 70(0) = 200 \]

### Step 8: Determine the Minimum Cost
The minimum cost occurs at the point \( (2, 4) \) with a cost of Rs. 380.

### Final Answer
Thus, the minimum cost of the mixture is Rs. 380.

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