To solve the problem of mixing wheat and rice to meet the minimum daily requirements of proteins and carbohydrates at the minimum cost, we can follow these steps:
### Step 1: Define Variables
Let:
- \( X \) = quantity of wheat (in grams)
- \( Y \) = quantity of rice (in grams)
### Step 2: Establish Cost Function
The cost of wheat is Rs 4 per kg (or Rs 0.004 per gram) and rice is Rs 6 per kg (or Rs 0.006 per gram). Therefore, the cost function \( Z \) can be expressed as:
\[
Z = 0.004X + 0.006Y
\]
### Step 3: Set Up Constraints
From the problem, we know the following nutritional values:
- Wheat provides 0.1 gm of protein and 0.25 gm of carbohydrates per gram.
- Rice provides 0.05 gm of protein and 0.5 gm of carbohydrates per gram.
The minimum daily requirements are:
- Protein: 50 gm
- Carbohydrates: 200 gm
We can set up the constraints based on these requirements:
1. Protein constraint:
\[
0.1X + 0.05Y \geq 50
\]
2. Carbohydrate constraint:
\[
0.25X + 0.5Y \geq 200
\]
### Step 4: Simplify Constraints
To simplify the constraints, we can multiply through by 100 to eliminate decimals:
1. Protein constraint:
\[
10X + 5Y \geq 5000 \quad \text{(divide by 5)} \quad 2X + Y \geq 1000 \quad \text{(Equation 1)}
\]
2. Carbohydrate constraint:
\[
25X + 50Y \geq 20000 \quad \text{(divide by 25)} \quad X + 2Y \geq 800 \quad \text{(Equation 2)}
\]
### Step 5: Find Intercepts for Graphing
To graph the constraints, we find the intercepts for each equation.
**For Equation 1: \( 2X + Y = 1000 \)**
- If \( X = 0 \): \( Y = 1000 \) (Point A)
- If \( Y = 0 \): \( X = 500 \) (Point B)
**For Equation 2: \( X + 2Y = 800 \)**
- If \( X = 0 \): \( Y = 400 \) (Point C)
- If \( Y = 0 \): \( X = 800 \) (Point D)
### Step 6: Plot the Graph
Plot the points A (0, 1000), B (500, 0), C (0, 400), and D (800, 0) on a graph. Draw the lines for both equations and identify the feasible region.
### Step 7: Find Corner Points
The corner points of the feasible region can be found by solving the equations simultaneously:
1. Solve \( 2X + Y = 1000 \) and \( X + 2Y = 800 \).
From \( 2X + Y = 1000 \):
\[
Y = 1000 - 2X
\]
Substituting into \( X + 2Y = 800 \):
\[
X + 2(1000 - 2X) = 800
\]
\[
X + 2000 - 4X = 800
\]
\[
-3X = -1200 \quad \Rightarrow \quad X = 400
\]
Substituting \( X = 400 \) back into \( Y = 1000 - 2(400) \):
\[
Y = 1000 - 800 = 200
\]
Thus, one corner point is (400, 200).
### Step 8: Evaluate Cost at Corner Points
Evaluate the cost function \( Z = 0.004X + 0.006Y \) at the corner points:
1. \( (0, 1000) \):
\[
Z = 0.004(0) + 0.006(1000) = 6
\]
2. \( (500, 0) \):
\[
Z = 0.004(500) + 0.006(0) = 2
\]
3. \( (400, 200) \):
\[
Z = 0.004(400) + 0.006(200) = 1.6 + 1.2 = 2.8
\]
### Step 9: Identify Minimum Cost
The minimum cost occurs at the point \( (400, 200) \) with a cost of Rs 2.8.
### Final Answer
To meet the minimum daily requirements of proteins and carbohydrates at minimum cost, the quantities should be:
- Wheat: 400 grams
- Rice: 200 grams
