To solve the problem using linear programming, we will follow these steps:
### Step 1: Define the Variables
Let:
- \( X \) = kilograms of fertilizer type A
- \( Y \) = kilograms of fertilizer type B
### Step 2: Define the Objective Function
We want to minimize the cost:
\[
Z = 5X + 8Y
\]
### Step 3: Define the Constraints
From the problem, we know that:
1. The nitrogen requirement:
\[
0.10X + 0.05Y \geq 7
\]
Simplifying this gives:
\[
2X + Y \geq 140 \quad \text{(1)}
\]
2. The phosphoric acid requirement:
\[
0.06X + 0.10Y \geq 7
\]
Simplifying this gives:
\[
3X + 5Y \geq 350 \quad \text{(2)}
\]
3. Non-negativity constraints:
\[
X \geq 0, \quad Y \geq 0
\]
### Step 4: Graph the Constraints
To graph the constraints, we will find the intercepts for each equation.
**For Equation (1): \( 2X + Y = 140 \)**
- When \( X = 0 \): \( Y = 140 \) (point (0, 140))
- When \( Y = 0 \): \( 2X = 140 \) → \( X = 70 \) (point (70, 0))
**For Equation (2): \( 3X + 5Y = 350 \)**
- When \( X = 0 \): \( 5Y = 350 \) → \( Y = 70 \) (point (0, 70))
- When \( Y = 0 \): \( 3X = 350 \) → \( X = \frac{350}{3} \approx 116.67 \) (point (116.67, 0))
### Step 5: Find the Points of Intersection
To find the intersection of the two lines, we solve the equations:
1. \( 2X + Y = 140 \)
2. \( 3X + 5Y = 350 \)
From (1), we can express \( Y \):
\[
Y = 140 - 2X
\]
Substituting into (2):
\[
3X + 5(140 - 2X) = 350
\]
\[
3X + 700 - 10X = 350
\]
\[
-7X + 700 = 350
\]
\[
-7X = -350
\]
\[
X = 50
\]
Substituting \( X = 50 \) back into \( Y = 140 - 2X \):
\[
Y = 140 - 2(50) = 40
\]
Thus, the intersection point is \( (50, 40) \).
### Step 6: Identify the Corner Points
The corner points of the feasible region are:
1. \( (0, 140) \)
2. \( (70, 0) \)
3. \( (0, 70) \)
4. \( (116.67, 0) \)
5. \( (50, 40) \)
### Step 7: Evaluate the Objective Function at Each Corner Point
1. \( (0, 140) \): \( Z = 5(0) + 8(140) = 1120 \)
2. \( (70, 0) \): \( Z = 5(70) + 8(0) = 350 \)
3. \( (0, 70) \): \( Z = 5(0) + 8(70) = 560 \)
4. \( (116.67, 0) \): \( Z = 5(116.67) + 8(0) \approx 583.35 \)
5. \( (50, 40) \): \( Z = 5(50) + 8(40) = 250 + 320 = 570 \)
### Step 8: Determine the Minimum Cost
The minimum cost occurs at the point \( (50, 40) \) with:
\[
Z = 570
\]
### Conclusion
To meet the requirements and minimize costs, the farmer should buy:
- 50 kg of fertilizer type A
- 40 kg of fertilizer type B
