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Question
A farmer has a supply of chemical fertilizer of type A which contains 10% nitrogen and 6% phosphoric acid and of type B which contains 5% nitrogen and 10% phosphoric acid. After the soil test, it is found that at least 7 kg of nitrogen and the same quantity of phosphoric acid is required for a good crop. The fertilizer of type A costs ₹ 5.00 per kg and the type B costs ₹ 8.00 per kg. Using Linear programming, find how many kilograms of each type of fertilizer should be bought to meet the requirement and for the cost to be minimum. Find the feasible region in the graph.
Solution
Let the farmer use x kg of fertilizer of type A and y kg of fertilizer of type B.
The L.P.P. as per given statement of the question is :
We have to minimize Z = 5x + 8y, subject to the constraints:
`(10)/(100) xx x + (5)/(100) xx y ≥7`
or 10x + 5y ≥ 700
2x + y ≥ 140 ...(i)
`(6)/(100) xx x + (10)/(100) xx y ≥ 7`
or 6x + 10y ≥ 700
3x + 5y ≥ 350
x ≥ 0, y ≥ 0 ...(ii)
Consider 2x + y = 140
table of the solution is :
| x | 70 | 0 |
| y | 0 | 140 |
Consider 3x + 5y = 350
Table of the solution is :
| x | 0 | 115 |
| y | 70 | 1 |
On plotting the graph of the above constraints or inequalities, we obtained the shaded region having corner points A, E, D as a feasible region.
Now, the value of Z is evaluated at corner points in the following table:
| Corner point | Z = 5x + 8y |
| A `(350/3 , 0)` | Z =5 x `(350)/(3)` + 0 = 583.33 |
| E(50, 40) | Z = 5 x 50 + 8 x 40 = 570 (Minimum) |
| D(0, 140) | Z = 0 + 8 x 140 = 1120 |
Since feasible region is unbounded.
∴ We have to draw the graph of the inequality 5x + 8y < 570.
The graph of the above inequality does not have any common point except (50, 40), so minimum value of Z = Rs 570 at (50, 40).
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