Let the farmer mixes x bags of brand P and y bags of brand Q. We construct the following table:
| Brands |
Number of bags |
Element A (units) |
Element B (units) |
Element C (units) |
Cost (per bag) |
| P |
x |
3x |
2.5x |
2x |
250x |
| Q |
y |
1.5y |
11.25y |
3y |
200y |
| Total |
x + y |
3x + 1.5y |
2.5x + 11.25y |
2x + 3y |
250x + 200y |
| Minimum Requires |
|
18 |
45 |
24 |
|
So, our problem is minimize Z = 250x + 200y …(i)
Subject to constraints 3x + 1.5y ≥ 18
⇒ 2x + y ≥ 12 …(ii)
2.5x + 11.25y ≥ 45
⇒ 2x + 9y ≥ 36 …(iii)
2x + 3y ≥ 24 …(iv)
x ≥ 0, y ≥ 0 …(v)
Firstly, draw the graph of the line 3x + 1.5y = 18
Secondly, draw the graph of the line 2.5x + 11.25y = 45
Thirdly, draw the graph of the line 2x + 3y = 24
On solving equations 3x + 1.5y = 18 and 2x + 3y = 24, we get C(3, 6).
Similarly, on solving equations 2.5x + 11.25y = 45 and2x + 3y = 24, we get B(9, 2).
The corner points of the feasible region are A(18, 0), B(9, 2), C(3, 6) and D (0, 12). (See below figure)
The values of Z at these points are as follows:
As the feasible region is unbounded, therefore 1950 may or may not be the minimum value of Z.
For this, we draw a graph of the inequality 250x + 200y < 1950 or
5x + 4y < 39 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 5x + 4y < 39.
Therefore, the minimum value of Z is 1950 at C(3, 6).
Thus, 3 bags of brand P and 6 bags of brand Q should be used in the mixture to minimize the cost to Rs. 1950.
