Question

# A cooperrative society of farmers has 50 hectare of land to grow two crops X and Y. The profit from crop X and Y per hectare are estimated as Rs 10,500 and Rs 9,000 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 litres of herbicide should be used in order 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 maximise the total profit of the society?

Solution

## Let x hectare of land be allocated to crop X and y hectare of land be allocated to crop Y ⇒x≥0,y≥0 Profit per hectare on crop X= Rs 10,500 Profit per hectare on crop Y= Rs 9,000 Therefore, total profit = Rs 10500x+9000y The mathematical formulation of the problem is as follows: Maximise Z=10500x+9000y Subject to constraints: x+y≤50⋯(1)20x+10y≤800⇒2x+y≤80⋯(2)x, y≥0⋯(3) Now we can plot the graph for the above constraints. Our feasible region is bounded with corner points O(0,0),A(40,0),B(30,20) & C(0,50). Corner Points Z=10500x+9000y O(0,0) 0 A(40,0) 420000 B(30,20) 495000←Maximum  C(0,50) 450000   Hence, the society will get maximum profit of Rs 4,95,000 by allocating 30 hectares for crop X and 20 hectares for crop Y.

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