a cooperative society of farmers has 50 hectoare of land to grow two crops x and y . the profits from the crops x and y per hectare are estimated as rs 10500 and rs 9000 respectively. to control weeds, a liquid herbicide has to be used for crops x and y at rates of 20 litres and 10 litres per hectare . further no more than 800 litre of herbicide should be used to protect fish and wild life using a pond which collects , drainage from this land . how much land should be allocated to each crop so as to max the total profit of the society ? also find max profit
Let the land allocated for crop A be x hectares and crop B be y hectares.
Maximum area of the land available for two crops is 50 hectares.
x+y ≤ 50
Liquid herbicide to be used for crops A and B are at the rate of 20 litres and 10 litres per hectare respectively. Maximum amount of herbicide to be used is 800 litres.
20x+10y≤ 800
2x+y≤ 80
The profits from crops A and B per hectare are Rs 10,500 and Rs 9,000 respectively.
Thus, total profit = Rs (10,500 x+ 9,000y) = Rs 1500 (7x+ 6y)
Therefore, the mathematical formulation of the given problem is
Maximize Z = 1500 (7x+ 6y) subject to the constraints
x+y≤ 50... (1)
2x+y≤ 80... (2)
x≥ 0... (3)
y≥ 0... (4)
The feasible region determined by constraints (1), (2), (3) and (4) is represented by the
shaded region in the following graph:
The corner points of the feasible region are O (0, 0), A (40, 0), B (30, 20) and C (0, 50).
The values of Z at these corner points are calculated as:
Corner | Z=1500(7x+6y) |
|
O(0,0) | 0 |
|
A(40,0) | 420000 |
|
B(30.20) | 495000 | Maximum |
C(0,50) | 420000 |
|
The maximum profit is at point B (30, 20).
Therefore, 30 hectares of land should be allocated for crop A and 20 hectares of land should be allocated for crop B
The maximum profit is Rs 495000.