Question

A cooperative society of farmers has 50 hectare of land to grow two crops X and Y. The profit 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 at to maximise the total profit of the society?

Answer 1

Let the land allocated for crop X be x hectares and crop Y be y hectares.

Maximum area of the land available for two crops is 50 hectares.

∴ x + y ≤ 50

Liquid herbicide to be used for crops X and Y are at the rate of 20 litres and 10 litres per hectare respectively. Maximum amount of herbicide to be used is 800 litres.

∴ 20x + 10y ≤ 800

The profits from crops X and Y per hectare are Rs 10,500 and Rs 9,000 respectively.

Thus, total profit = Rs (10,500x + 9,000y)

Therefore, the mathematical formulation of the given problem is

Maximize Z = 10,500x + 9,000y subject to the constraints

x + y ≤ 50

2x + y ≤ 80

x ≥ 0

y ≥ 0

First we will convert inequations into equations as follows:
x + y = 50, 2x + y = 80, x = 0 and y = 0

Region represented by x + y ≤ 50:
The line x + y = 50 meets the coordinate axes at A(50, 0) and $B\left(0,50\right)$ respectively. By joining these points we obtain the line
x + y = 50. Clearly (0,0) satisfies the x + y = 50. So, the region which contains the origin represents the solution set of the inequation
x + y ≤ 50.

Region represented by 20x + 10y ≤ 800:
The line 20x + 10y = 800 meets the coordinate axes at C(40, 0) and $D\left(0,80\right)$ respectively. By joining these points we obtain the line
20x + 10y = 800. Clearly (0,0) satisfies the inequation 20x + 10y ≤ 800. So,the region which contains the origin represents the solution set of the inequation 20x + 10y ≤ 800.

Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints x + y ≤ 50, 20x + 10y ≤ 800, x ≥ 0 and y ≥ 0 are as follows.

The corner points of the feasible region are O(0, 0), C(40, 0), E(30, 20) and B(0, 50).

The values of Z at these corner points are calculated as:

Corner point	Z = 10,500x + 9,000y
O(0, 0)	0
C(40, 0)	4,20,000
E(30, 20)	4,95,000	Maximum
B(0, 50)	4,50,000

The maximum profit is at point E(30, 20).

Therefore, 30 hectares of land should be allocated for crop X and 20 hectares of land should be allocated for crop Y.

The maximum profit is Rs 4,95,000.