| LP Problem |
| Ha of Rice, Wheat and Cotton to be irrigated 100%, 80% and 60% to maximize profit |
| Subject to: Available land, and water |
It can be shown as follows:
| Maximize |
| 190x(9R1+8.28R2+7.02R3 |
| 180x(12W1+10.56W2+9.6W3 |
| 240x(8C1+5.76C2+5.52C3 |
Where R1, R2, R3 is Ha of Rice irrigated at 100%, 80% and 60% respectively
Where W1, W2, W3 is Ha of Wheat irrigated at 100%, 80% and 60% respectively
Where C1, C2, C3 is Ha of Cotton irrigated at 100%, 80% and 60% respectively
Subject to
Land use
R1+R2+R3 = 30
W1+W2+W3 = 40
C1+C2+C3 = 30
Water use
1300(R1+0.8R2+0.6R3) + 1200(W1+0.8W2+0.6W3) + 1500(C1+0.8C2+0.6C3) <=125000
This is the solution we get in Excel, using Solver:
3.08 Ha of Rice at 100%, and 26.92 Ha of Rice at 80% irrigation
40 Ha of Wheat at 100% irrigation
30 Ha of Cotton at 100% irrigation
give $191,617 of profit
| Land Used | Production in tons / Ha | Profit ($/ton) | ||||||||||||
| Land used (ha) | Water (m3/ha) | Water demand @100% | Irrigation @100% | Irrigation @80% | Irrigation @60% | Subject to | Irrigation @100% | Irrigation @80% | Irrigation @60% | Profit | Water | |||
| Rice | 30 | 1300 | 39000 | 3.08 | 26.92 | 0 | 30 | 9 | 8.28 | 7.02 | 190 | 47,617 | 32000 | 30 |
| Wheat | 40 | 1200 | 48000 | 40.00 | - | 0 | 40 | 12 | 10.56 | 9.6 | 180 | 86,400 | 48000 | 40 |
| Cotton | 30 | 1500 | 45000 | 30.00 | - | 0 | 30 | 8 | 5.76 | 5.52 | 240 | 57,600 | 45000 | 30 |
| Total required | 100 | 132000 | 1,91,617 | 125000 | ||||||||||
| Total available | 100 | 125000 | ||||||||||||
Problem 4 (24 points) Farmer Joe is planning to plant 30 hectares of rice, 40 hectares...