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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:

12100x+4100y1212x+4y12003x+y300 .....1
Also,

5x100+5y100125x+5y1200x+y240 .....2
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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