Answer:
Let the land allocated for crop A be x hectare and crop B be y hectare. Then, maximum area of the land available for two crops is 50 hectare. \[x+y\le 50\] Liquid herbicide to be used for crops A and B are at the rate of 20 L and 10 L per hectare, respectively. Maximum amount of herbicide to be used is 800 L. \[20x+10y\le 800\] and \[2x+y\le 80\] The profits from crops A and B per hectare are Rs. 10500 and Rs. 9000 respectively. Thus, total profit \[=(10500x+9000y)=Rs.\,1500\,\,(7x+6y)\] The mathematical formulation of the given problem is Maximise\[Z=1500\,\,(7x+6y)\] Subject to the constraints \[x+y\le 50\] ?(i) \[2x+y\le 80\] ?(ii) and \[x,\,\,y\ge 0\] ?(iii) The corner points of 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
The maximum profit is at point B (30, 20) \[\therefore \]30 hec of land should be allocated for crop A and 20 hec of land should be allocated for crop B. Value Yes, because excess use of herbicide can make drainage water poisonous and shows it harms the life of water living creature and wild life. Corner \[Z=1500\,\,(7x+6y)\] \[O\,(0,\,\,0)\] \[Z=0\] \[A\,(40,\,\,0)\] \[Z=1500\,(7\times 40+6\times 0)=420000\] \[B\,(30,\,\,20)\] \[Z=1500\,(7\times 30+6\times 20)\] \[=49500\,\,(maximum)\] \[C\,(0,\,\,50)\] \[Z=1500\,(7\times 0+6\times 50)=420000\]
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