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Question
There are two types of fertilizers F1 and F2. F1 consists of 10% nitrogen and 6% phosphoric acid and F2 consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer finds that she needs at least 14 kg of nitrogen and 14 kg of phosphoric acid for her crop. If F1 cost Rs 6/kg and F2 costs Rs 5/kg, determine how much of each type of fertilizer should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?
Solution
Let the farmer buy x kg of fertilizer F1 and y kg of fertilizer F2. T`herefore,
x ≥ 0 and y ≥ 0
The given information can be complied in a table as follows.
| Nitrogen (%) | Phosphoric Acid (%) | Cost (Rs/kg) | |
| F1 (x) | 10 | 6 | 6 |
| F2 (y) | 5 | 10 | 5 |
|
Requirement (kg) |
14 | 14 |
F1 consists of 10% nitrogen and F2 consists of 5% nitrogen. However, the farmer requires at least 14 kg of nitrogen.
∴ 10% of x + 5% of y ≥ 14
`x/10+y/20 >=14`
`2x+y>=280`
F1 consists of 6% phosphoric acid and F2 consists of 10% phosphoric acid. However, the farmer requires at least 14 kg of phosphoric acid.
∴ 6% of x + 10% of y ≥ 14
6x/100 + 10y/100 >=14
3x +56y >= 700
Total cost of fertilizers, Z = 6x + 5y
The mathematical formulation of the given problem is
Minimize Z = 6x + 5y … (1)
subject to the constraints,
2x + y ≥ 280 … (2)
3x + 5y ≥ 700 … (3)
x, y ≥ 0 … (4)
The feasible region determined by the system of constraints is as follows.
It can be seen that the feasible region is unbounded.
The corner points are A(700/3,0),B(100,80) and C(0,280)
The values of Z at these points are as follows.
|
Corner point |
Z = 6x + 5y | |
| A(700/3,0) | 1400 | |
| B(100, 80) | 1000 | → Minimum |
| C(0, 280) | 1400 |
As the feasible region is unbounded, therefore, 1000 may or may not be the minimum value of Z.
For this, we draw a graph of the inequality, 6x + 5y < 1000, and check whether the resulting half plane has points in common with the feasible region or not.
It can be seen that the feasible region has no common point with
6x + 5y < 1000
Therefore, 100 kg of fertiliser F1 and 80 kg of fertilizer F2 should be used to minimize the cost. The minimum cost is Rs 1000.
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