Question -
Answer -
Let the farmer buy x kg of fertilizer F1 andy kg of fertilizer F2.Hence,
x ≥ 0 and y ≥ 0
The given information can be compiled in a table is given below
| Nitrogen (%) | Phosphoric Acid (%) | Cost (Rs / kg) |
| F1 (x) | 10 | 6 | 6 |
| F2 (y) | 5 | 10 | 5 |
| Requirement (kg) | 14 | 14 | |
F1 consistsof 10% nitrogen and F2 consistsof 5% nitrogen.
However, the farmer requires at least 14 kg of nitrogen
So, 10% of x + 5% of y ≥ 14
x / 10 + y / 20 ≥ 14
By L.C.M we get
2x + y ≥ 280
F1 consistsof 6% phosphoric acid and F2 consists of 10% phosphoric acid.
However, the farmer requires at least 14 kg of phosphoric acid
So, 6% of x + 10 % of y ≥ 14
6x / 100 + 10y / 100 ≥ 14
3x + 5y ≥ 700
Total cost of fertilizers, Z = 6x + 5y
The mathematical formulation of the given problem can be writtenas
Minimize Z = 6x + 5y ………….. (i)
Subject to the constraints,
2x + y ≥ 280 ……… (ii)
3x + 5y ≥ 700 ………. (iii)
x, y ≥ 0 …………. (iv)
The feasible region determined by the system of constraints isgiven below
Here, we can see that the feasible region is unbounded.
A (700 / 3, 0), B (100, 80) and C (0, 280) are the corner points
The values of Z at these points are given below
| Corner point | Z = 6x + 5y | |
| A (700 / 3, 0) | 1400 | |
| B (100, 80) | 1000 | Minimum |
| C (0, 280) | 1400 | |
Here, the feasible region is unbounded, hence, 1000 may or maynot be the minimum value of Z.
For this purpose, we draw a graph of the inequality, 6x + 5y< 1000, and check whether the resulting half plane has points in common withthe feasible region or not.
Here, it can be seen that the feasible region has no commonpoint with 6x + 5y < 1000
Hence, 100 kg of fertilizer F1 and 80 kg of fertilizer F2 should be used to minimize the cost. The minimum cost isRs 1000