Answer to Question #283396 in Operations Research for Andinet

Question

11. Wodera 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 Birr 10,500 and Birr 9,000 respectively. To control weeds, a liquid herbicide has to be used for crops X and Y at rates of 20 liters and 10 liters per hectare. Further, no more than 800 liters 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 maximize the total profit of the society? (Solve By graphic method).

Answer 1

profit:

$Z=10500x_1+9000x_2$

total number of hectares:

$x_1+x_2\le50$

amount of of herbicide:

$20x_1+10x_1\le 800$

x₁ is number of hectares for crop X,

x₂ is number of hectares for crop Y

for Extreme Points:

$x_1+x_2=50$

$x_1=0\implies x_2=50$

$20x_1+10x_2=800$

$x_2=0\implies x_1=40$

for intersection of graphs:

$x_1=50-x_2$

$2(50-x_2)+x_2=80$

$x_2=20,x_1=30$

Objective function values at extreme points:

$Z(40,0)=420000$

$Z(0,50)=450000$

$Z(30,20)=495000$

The maximum value of the objective function Z=495000 occurs at the extreme point (30,20).

Hence, the optimal solution to the given LP problem is : x₁=30, x₂=20 and max Z=495000