There are two types of fertilizers `F_(1)` and `F_(2).F_(1)` consists of 10% nitrogen and 6% phosphoric acid and `F_(2)` contains 5% nitrogen and 10% phosphoric acid. After testing the soil condition a farmer finds that he needs at least 14 kg of nitrogen and 14 kg of phosphoric acid forhis crop. If `F_(1)` costs Rs. 6 per kg and `F_(2)` costs Rs. 5 perkg, determine how muchofeach of fertilizers beused so that nutrient requirements are met at a minimum cost. What is the minimum cost?

Answer

To solve the problem, we need to set up a system of equations based on the information provided about the fertilizers and the requirements for nitrogen and phosphoric acid. Let's break this down step by step.

### Step 1: Define Variables
Let:
- \( x \) = amount (in kg) of fertilizer \( F_1 \)
- \( y \) = amount (in kg) of fertilizer \( F_2 \)

### Step 2: Write Down the Nutrient Content
From the problem, we know:
- Fertilizer \( F_1 \) contains:
- 10% nitrogen
- 6% phosphoric acid

- Fertilizer \( F_2 \) contains:
- 5% nitrogen
- 10% phosphoric acid

### Step 3: Set Up the Constraints
The farmer needs at least:
- 14 kg of nitrogen
- 14 kg of phosphoric acid

From the percentages, we can express the total nitrogen and phosphoric acid from both fertilizers:
1. For nitrogen:
\[ 0.10x + 0.05y \geq 14 \]
Multiplying through by 100 to eliminate decimals:
\[ 10x + 5y \geq 1400 \quad \text{(Constraint 1)} \]

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

### Step 4: Set Up the Objective Function
The cost function to minimize is given by:
\[ Z = 6x + 5y \]

### Step 5: Solve the Constraints
We will rewrite the constraints in standard form:
1. \( 10x + 5y \geq 1400 \)
2. \( 6x + 10y \geq 1400 \)

### Step 6: Find the Intercepts
To graph the constraints, we can find the intercepts for each equation.

**For Constraint 1:**
- Let \( x = 0 \):
\[ 5y = 1400 \Rightarrow y = 280 \quad (0, 280) \]
- Let \( y = 0 \):
\[ 10x = 1400 \Rightarrow x = 140 \quad (140, 0) \]

**For Constraint 2:**
- Let \( x = 0 \):
\[ 10y = 1400 \Rightarrow y = 140 \quad (0, 140) \]
- Let \( y = 0 \):
\[ 6x = 1400 \Rightarrow x = \frac{1400}{6} \approx 233.33 \quad \left(\frac{700}{3}, 0\right) \]

### Step 7: Graph the Constraints
Plot the points \( (0, 280) \), \( (140, 0) \), \( (0, 140) \), and \( \left(\frac{700}{3}, 0\right) \) on a graph. Draw the lines for each constraint and identify the feasible region.

### Step 8: Identify Corner Points
The corner points of the feasible region will be where the constraints intersect. We can find the intersection points by solving the equations:
1. \( 10x + 5y = 1400 \)
2. \( 6x + 10y = 1400 \)

### Step 9: Solve the System of Equations
From the first equation:
\[ y = \frac{1400 - 10x}{5} = 280 - 2x \]

Substituting into the second equation:
\[ 6x + 10(280 - 2x) = 1400 \]
\[ 6x + 2800 - 20x = 1400 \]
\[ -14x = -1400 \Rightarrow x = 100 \]
Substituting \( x = 100 \) back into \( y = 280 - 2(100) \):
\[ y = 80 \]

### Step 10: Calculate the Minimum Cost
Now we can find the minimum cost using the values of \( x \) and \( y \):
\[ Z = 6(100) + 5(80) = 600 + 400 = 1000 \]

### Final Answer
The farmer should use:
- \( 100 \) kg of fertilizer \( F_1 \)
- \( 80 \) kg of fertilizer \( F_2 \)

The minimum cost is Rs. \( 1000 \).