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  • An oil company has two depots A and B with capacities of 7000 L and 4000 L respectively. The company is to supply oil to three petrol pumps, D, E and F whose requirements are 4500L, 3000L and 3500L respectively.

7.An oil company has two depots A and B with capacities of 7000 L and 4000 L respectively. The company is to supply oil to three petrol pumps, D, E and F whose requirements are 4500L, 3000L and 3500L respectively. The distances (in km) between the depots and the petrol pumps is given in the following table:

Distance in (km.)
From/To A B
D 7 3
E 6 4
F 3 2

Assuming that the transportation cost of 10 litres of oil is Re 1 per km, how should the delivery be scheduled in order that the transportation cost is minimum? What is the minimum cost?

Answers (1)

best_answer

Let   x and y litres of oil be supplied from A to petrol pump,D and E. Then , (7000-x-y) will be  supplied from A to petrol pump F.

Requirements at petrol pump  D is  4500 L. since   x L A are  transported from depot A,remaining 4500-x L will be    transported from petrol pump  B

Similarly, (3000-y)L  and 3500-(7000-x-y)=(x+y-3500) L will be transported from depot  B to petrol E and F respectively.

The problem can be represented diagrammatically as follows:

   

          x,y\geq 0  and  7000-x-y\geq 0  

          x,y\geq 0  and  x+y\leq 7000

 

4500-x\geq 0,3000-y\geq 0\, \, \, and\, \, x+y-3500\geq 0

 \Rightarrow \, \, \, \, x\leq 4500,y\leq 3000,x+y\geq 3500

Cost of transporting 10 L petrol =Re 1

Cost of transporting 1 L petrol =\frac{1}{10}

Total transportation cost z is given by , 

z=\frac{7}{10}x+\frac{6}{10}y+\frac{3}{10}(7000-x-y)+\frac{3}{10}(4500-x)+\frac{4}{10}(3000-y)+\frac{2}{10}(x+y-3500)

z=0.3x+0.1y+3950

  Mathematical formulation of given problem is as  follows:

Minimize : z=0.3x+0.1y+3950

Subject to constraint ,

                           x+y\leq 7000

                           x\leq 4500

                        y\leq 3000

                         x+y\geq 3500

                            x,y\geq 0

The feasible  region determined by constraints is as follows:

       

The corner points of feasible region are A(3500,0),B(4500,0),C(4500,2500),D(4000,3000),E(500,3000)

The value of Z at corner points is as shown :

 corner points 

z=0.3x+0.1y+3950

  
   A(3500,0)             5000  

B(4500,0)

              5300  
  C(4500,2500)               5550  
E(500,3000)             4400 minimum
   D(4000,3000)               5450

 

Hence , Z has miniimum value  4400  at  point  E(500,3000)

Posted by

seema garhwal

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