A cooperative society of farmers has 50 hectare of land to grow two crops X and Y. The profits from crops 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?

Let x hectare of land be allocated to crop X and y hectare to crop Y. Clearly, x ≥ 0, y ≥ 0.
Profit per hectare on crop X = Rs. 10500
Profit per hectare on crop Y = Rs. 9000
∴  total profit = Rs. (10500 x + 9000 y)
The mathematical formulation of the problem is as follows:
Maximise    Z = 10500 x + 9000 y
subject to constraints x + y ≤ 50, 20x + 10y ≤ 800
i.e. 2 x + y ≤ 80
and    x ≥ 0, y ≥ 0
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Now we draw the graph of x + y = 50
For x = 0, y = 50
For y = 0, x = 50
∴ line meets OX in A(50, 0) and OY in L(0, 50).
Again we draw the graph of 2x + y = 80.
For x = 0, y = 80
For y = 0, 2x = 80 or x = 40

∴  line meets OX in B(40, 0) and OY in M(0, 80).
Since feasible region satisfies all the constraints.
∴ OBCL is the feasible region.
The corner points are O(0, 0), B(40, 0), C(30, 20), L(0, 50).
At O(0, 0), Z = 0 + 0 = 0
At B(40, 0), Z = 420000 + 0 = 420000
At C(30, 20), Z = 315000 + 180000 = 495000
At L(0, 50), Z = 0 + 450000 = 450000
∴ maximum value = 495000 at (30, 20)
∴ society will get the maximum profit of Rs. 495000 by allocating 30 hectares for crop X and 20 hectares for crop Y.