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Question
There are two types of fertilisers 'A' and 'B' . 'A' consists of 12% nitrogen and 5% phosphoric acid whereas 'B' consists of 4% nitrogen and 5% phosphoric acid. After testing the soil conditions, farmer finds that he needs at least 12 kg of nitrogen and 12 kg of phosphoric acid for his crops. If 'A' costs ₹10 per kg and 'B' cost ₹8 per kg, then graphically determine how much of each type of fertiliser should be used so that nutrient requiremnets are met at a minimum cost
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Solution
The given information can tabulated as follows:
| Fertilizer | Nitrogen | Phosphoric Acid | Cost/kg (in ₹) |
| A | 12% | 5% | 10 |
| B | 4% | 5% | 8 |
Let the requirement of fertilizer A by the farmer be x kg and that of B be y kg.
It is given that farmer requires atleast 12 kg of nitrogen and 12 kg of phosphoric acid for his crops.
The inequations thus formed based on the given information are as follows:
\[\frac{12}{100}x + \frac{4}{100}y \geq 12\]
\[ \Rightarrow 12x + 4y \geq 1200\]
\[ \Rightarrow 3x + y \geq 300 . . . . . \left( 1 \right)\]
Also,
\[\frac{5x}{100} + \frac{5y}{100} \geq 12\]
\[ \Rightarrow 5x + 5y \geq 1200\]
\[ \Rightarrow x + y \geq 240 . . . . . \left( 2 \right)\]
Total cost of the fertilizer Z = ₹ (10x + 8y)
Therefore, the mathematical formulation of the given linear programming problem can be stated as:
Minimize Z = 10x + 8y
Subject to the constraints
3x + y ≥ 300 .....(1)
x + y ≥ 240 .....(2)
x ≥ 0, y ≥ 0 .....(3)
The feasible region determined by constraints (1) to (3) is graphically represented as:
Here, it is seen that the feasible region is unbounded. The values of Z at the corner points of the feasible region are represented in tabular form as:
| Corner Point | Z = 10x + 8y |
| A(0, 300) | Z = 10 × 0 + 8 × 300 = 2400 |
| B(30, 210) | Z = 10 × 30 + 8 × 210 = 1980 |
| C(240, 0) | Z = 10 × 240 + 8 × 0 = 2400 |
The open half plane determined by 10x + 8y < 1980 has no point in common with the feasible region. So, the minimum value of Z is 1980.
The minimum value of Z is 1980, which is obtained at x = 30 and y = 210.
Thus, the minimum requirement of fertilizer of type A will be 30 kg and that of type B will be 210 kg.
Also, the total minimum cost of the fertilisers is ₹ 1980.
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