CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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
x0,y0
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+y50(1)20x+10y8002x+y80(2)x, y0(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) 495000Maximum 
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.

flag
 Suggest corrections
thumbs-up
 
0 Upvotes



footer-image