(Allocation Problem)A farmer has a supply of chemical fertilizer and type I which contains 10% nitrogen and 6% phospheric acid and type II fertilizer which contains 5% nitrogen and 10% phospheric acid. After testing the soil condition of a field, it is found that at least 14 kg of nitrogen and 14kg of phospheric acid are required for a good crop.The fertilizer type I costs ₹.2.00 per kg and the type II ₹ 3.00 per kg.How many kilograms of each fertilizer should be used to meet the requirements and the cost be minimum?

Answer

To solve the allocation problem involving the farmer's fertilizer requirements, we will follow these steps:

### Step 1: Define Variables
Let:
- \( x \) = kilograms of Type I fertilizer
- \( y \) = kilograms of Type II fertilizer

### Step 2: Set Up the Constraints
From the problem, we know:
- Type I fertilizer contains 10% nitrogen and 6% phosphoric acid.
- Type II fertilizer contains 5% nitrogen and 10% phosphoric acid.

The requirements for the crop are:
- At least 14 kg of nitrogen
- At least 14 kg of phosphoric acid

We can set up the following inequalities based on the nutrient content:
1. For nitrogen:
\[ 0.10x + 0.05y \geq 14 \]
Multiplying through by 100 to eliminate decimals:
\[ 10x + 5y \geq 1400 \quad \text{(1)} \]

2. For phosphoric acid:
\[ 0.06x + 0.10y \geq 14 \]
Multiplying through by 100:
\[ 6x + 10y \geq 1400 \quad \text{(2)} \]

### Step 3: Set Up the Objective Function
The cost of the fertilizers is given as:
- Type I: ₹2.00 per kg
- Type II: ₹3.00 per kg

Thus, the objective function to minimize the cost is:
\[ Z = 2x + 3y \]

### Step 4: Solve the Constraints
Now we will rearrange the inequalities for graphing:

1. From (1):
\[ 2x + y \geq 280 \quad \text{(divide by 5)} \]

2. From (2):
\[ 3x + 5y \geq 700 \quad \text{(divide by 2)} \]

### Step 5: Find Intercepts for Graphing
For the first constraint \( 2x + y = 280 \):
- When \( x = 0 \), \( y = 280 \) (y-intercept)
- When \( y = 0 \), \( x = 140 \) (x-intercept)

For the second constraint \( 3x + 5y = 700 \):
- When \( x = 0 \), \( y = 140 \) (y-intercept)
- When \( y = 0 \), \( x = \frac{700}{3} \approx 233.33 \) (x-intercept)

### Step 6: Graph the Constraints
Plot the lines on a graph with the x-axis and y-axis. The feasible region will be the area that satisfies both inequalities.

### Step 7: Identify Corner Points
The corner points of the feasible region can be found by solving the equations of the lines:
1. Solve \( 2x + y = 280 \) and \( 3x + 5y = 700 \):
- From \( 2x + y = 280 \), express \( y \):
\[ y = 280 - 2x \]
- Substitute into the second equation:
\[ 3x + 5(280 - 2x) = 700 \]
\[ 3x + 1400 - 10x = 700 \]
\[ -7x = -700 \implies x = 100 \]
- Substitute \( x = 100 \) back to find \( y \):
\[ y = 280 - 2(100) = 80 \]
- So one corner point is \( (100, 80) \).

2. The other corner points are \( (0, 280) \) and \( (233.33, 0) \).

### Step 8: Evaluate the Objective Function at Each Corner Point
1. At \( (0, 280) \):
\[ Z = 2(0) + 3(280) = 840 \]

2. At \( (100, 80) \):
\[ Z = 2(100) + 3(80) = 200 + 240 = 440 \]

3. At \( (233.33, 0) \):
\[ Z = 2(233.33) + 3(0) \approx 466.67 \]

### Step 9: Determine the Minimum Cost
The minimum cost occurs at the point \( (100, 80) \) with a cost of ₹440.

### Conclusion
The farmer should use:
- 100 kg of Type I fertilizer
- 80 kg of Type II fertilizer