Answer to Question #283396 in Operations Research for Andinet

Question #283396

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).

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Expert's answer
2021-12-31T13:05:38-0500

profit:

Z=10500x1+9000x2Z=10500x_1+9000x_2

total number of hectares:

x1+x250x_1+x_2\le50

amount of of herbicide:

20x1+10x180020x_1+10x_1\le 800

x1 is number of hectares for crop X,

x2 is number of hectares for crop Y




for Extreme Points:

x1+x2=50x_1+x_2=50

x1=0    x2=50x_1=0\implies x_2=50


20x1+10x2=80020x_1+10x_2=800

x2=0    x1=40x_2=0\implies x_1=40


for intersection of graphs:

x1=50x2x_1=50-x_2

2(50x2)+x2=802(50-x_2)+x_2=80

x2=20,x1=30x_2=20,x_1=30


Objective function values at extreme points:

Z(40,0)=420000Z(40,0)=420000

Z(0,50)=450000Z(0,50)=450000

Z(30,20)=495000Z(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


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