 Hello, and welcome to the session. I am Deepika here. Let's discuss the question which says An oil company has two depots A and B with capacities of 7000 liters and 4000 liters respectively. The company is to supply oil to three petrol pumps B, E and F whose requirements are 4500 liters, 3000 liters and 3500 liters respectively. The distances between the depots and the petrol pumps is given in the following table. Now the distance from petrol pump B to the depot A is 7 kilometer and to the depot B it is 3 kilometer. The distance of the petrol pump E to the depot A is 6 kilometer and to the depot B it is 4 kilometer. Again the distance from the petrol pump F to the depot A is 3 kilometer and to the depot B it is 2 kilometer. Assuming that the transportation cost of 10 liters of oil is rupee 1 per kilometer, how should the delivery be scheduled in order that the transportation cost is minimum, what is the minimum cost? Let's start the solution. Now according to the question an oil company has two depots A and B which supply oil to three petrol pumps G, E and F. So let X liters and Y liters of the oil be supplied from depot at A to petrol pumps D, E respectively. The given problem can be explained diagrammatically. Now here A and B are two depots which supply oil to three petrol pumps D, E and F. Now according to the given question the capacity of depot A is 7000 liters and the company is to supply oil to three petrol pumps D, E and F whose requirements are 4500 liters, 3000 liters and 3500 liters respectively. Since the capacity of depot A is 7000 liters and we have assumed that X liters and Y liters of the oil be supplied from depot at A to petrol pumps D and E respectively. Therefore 7000 minus X minus Y liters of oil can be supplied from depot A to petrol pump D which can be supplied from depot A to petrol pump F is 7000 minus X minus Y. Hence we have X greater than equal to 0, Y greater than equal to 0 and 7000 minus X minus Y greater than equal to 0. That is X greater than equal to 0, Y greater than equal to 0 and X plus Y less than equal to 7000. Again we are given the requirement of petrol pump D is 4500 liters and the requirement of petrol pump E is 3000 liters and requirement of petrol pump F is 3500 liters. X liters of oil has already been supplied from depot A therefore the remaining 4500 minus X liters of oil will be supplied from depot B. Similarly Y liters of oil has already been supplied from depot A therefore the remaining 3000 minus Y liters of oil will be supplied from depot B. Now again we are given the capacity of depot B is 4000 liters so we are given the capacity of depot B is 4000 liters since 4500 minus X liters of oil will be supplied from depot B to the petrol pump D and 3000 minus Y liters of oil will be supplied from depot B to petrol pump E therefore 4000 minus 4500 minus X plus 3000 minus Y liters of oil will be supplied from depot B to petrol pump F. Since requirements of oil at petrol pumps B is 4500 liters, 3000 liters, 3500 liters respectively. So quantity of oil supplied from depot B to petrol pumps DER 4500 minus X liters 3000 minus Y liters 4000 minus 4500 minus X. plus 3000 minus Y liters respectively so we have 4500 minus X greater than equal to 0 and 3000 minus Y greater than equal to 0 and 4000 minus 4500 minus X plus 3000 minus Y greater than equal to 0 now 4500 minus X greater than equal to 0 implies X less than equal to 4500 again 3000 minus Y greater than equal to 0 implies Y less than equal to 3000 now on solving this inequality we have X plus Y greater than equal to 3500 again we are given the transportation cost of 10 liters of oil is rupee 1 per kilometer that is we have the transportation cost 10 liters of oil is rupee 1 per kilometer or we can say the transportation cost of 1 liter of oil is rupee 0.1 per kilometer so we have the total cost is equal to the total distance traveled in kilometer from two depots to the three petrol pumps into rupee 0.1 now we are given the distance of petrol pump D from the depot A is 7 kilometer and from depot B it is 3 kilometer again the distance of E from A is 6 kilometer and from B is 4 kilometer again the distance of petrol pump F from depot A is 3 kilometer and from depot B it is 2 kilometer therefore the total transportation cost Z is given by Z is equal to 0.1 into 7X plus 65 plus 3 into 7000 minus X minus 5 plus 3 into 4500 minus X 4500 minus x plus 4 into 3000 minus y plus into x plus y minus 3500. This is equal to 0.1 into 7 x plus 6 y plus 21000 minus 3 x minus 3 y plus 13500 minus 3 3x plus 12000 minus 4y plus 2x plus 2y minus 7000 and this is again equal to 0.1 into 3x plus y plus 39500 and this is again equal to 0.3x 0.1y plus 3950. According to the question we have to find how should the delivery be scheduled in order that the transportation cost is minimum and we have to find the minimum cost. Hence the given problem reduces to minimize z is equal to 0.3x plus 0.15 plus 3950 subject to the constraints x plus y less than equal to 7000 let us give this as number 1 x plus y greater than equal to 3500 let us give this as number 2 x less than equal to 4500 let us give this as number 3 again y less than equal to 3000 now let us give this as number 4 and x greater than equal to 0 y greater than equal to 0 let us give these non-negative constraints as number 5. So z is equal to 0.3x plus 0.15 plus 3950 is our objective function now these are the constraints so we will draw the graph and find the feasible region subject to these given constraints. So first we will draw the line representing the equation x plus y is equal to 7000 corresponding to the inequality x plus y less than equal to 7000 now clearly the points 0 7000 and 7000 0 lie on the line x plus y is equal to 7000 therefore the graph of the line can be drawn by plotting points 0 7000 and 7000 0 and then joining them now this is a point 0 7000 and this is a point 7000 0 now this line represents the equation x plus y is equal to 7000 now this line divides the plane into two half planes so we will consider the half plane which will satisfy one now clearly origin satisfy this inequality so the half plane containing the origin is the graph of one now the equation corresponding to the inequality x plus y greater than equal to 3500 is x plus y is equal to 3500 clearly the points 0 3500 and 3500 0 satisfy the equation x plus y is equal to 3500 so we will plot these points on the same graph and then join them to get the equation of the line x plus y is equal to 3500 now this is the point 0 3500 and this is the point 3500 0 now this line divides the plane into two half planes so we will consider the half plane which will satisfy two now clearly the origin does not satisfy this inequality so the half plane which does not contain the origin is the graph of two that is we will consider the half plane which will not contain the origin again the equation corresponding to the inequality x less than equal to 4500 is x is equal to 4500 now it represents a line parallel to y axis on the graph which passes through 4500 0 so now we will draw this line on the same graph so this is the line representing the equation x is equal to 4500 again this line divides the plane into two half planes so we will consider the half plane which will satisfy three clearly the origin satisfy this inequality so we will consider the half plane which will contain the origin in a similar way we will represent y less than equal to 3000 graphically by drawing the line y is equal to 3000 now this is a line parallel to x axis passing through the point 0 3000 now this is the point 0 3000 again this line divides the plane into two half planes so we will consider the half plane which will satisfy four so clearly the half plane which contains the origin will satisfy four so we will consider the half plane which contains the origin again x greater than equal to 0 and y greater than equal to 0 implies the graph lies in the first quadrant only so by making use of the constraints 1 to 5 we get the shaded region a b b e as a feasible region that is the shaded region a b c g e represented by the constraints 1 to 5 is the feasible region now here we observe that the feasible region is bounded now the coordinates of these corner points are 500 3000 that is the coordinates of the coordinates of the point a are 500 3000 now the coordinates of point b are 3500 0 and the coordinates of the point c are 4500 and 0 the coordinates of b are 4500 and 2500 again the coordinates of point e are 4000 3000 that is here the feasible region is bounded with coordinates of corner points as a with coordinates 500 3000 b with coordinates 3500 0 c with coordinates 4500 0 and b with coordinates 4500 2500 and e with coordinates 4000 3000 now according to corner point method cost of transportation z will be minimum at any one of these points so we will evaluate z is equal to 0.3 x plus 0.1 y plus 3950 at each of these points so at the point a with coordinates 500 3000 z is equal to 0.3 x 500 plus 0.1 x 3000 plus 3950 and this is equal to 150 plus 300 plus 3950 and this is again equal to 4400 now at the point 3500 0 z is equal to 0.3 x 3500 plus 0.1 x 0 plus 3950 and this is equal to 1050 plus 0 plus 3950 and this is again equal to 5000 now 4500 0 z is equal to 0.3 x 4500 plus 0.1 x 0 plus 3950 and this is equal to 1350 plus 0 plus 3950 and this is again equal to 3500 now at the point 4500 2500 z is equal to 0.3 x 4500 plus 0.1 x 2500 plus 3950 and this is equal to 1350 plus 250 plus 3950 and this is equal to 5550 now again at the point 4000 3000 z is equal to 0.3 x 4000 plus 0.1 x 3000 plus 3950 and this is equal to 1200 plus 300 plus 3950 and this is again equal to 5450 so here we observe that the transportation cost z is minimum at 500 3000 and its value at this point is 4400 hence the minimum value of z is 4400 at the point 500 3000 hence the delivery we should do in such a way that from depot A quantity of oil supplied to petrol pumps D should be 500 liters 3000 liters 3500 liters respectively quantity of oil to pumps from depot B should be 4000 liters 0 liters 0 liters respectively so by this way the transportation cost will be minimum and the minimum transportation cost will be rupees 4400 hence the answer for the question is from depot A 500 3000 and 3500 liters and from depot B 4000 0 and 0 liters should be supplied to petrol pumps D, E and F respectively and the minimum cost by this way will be rupees 4400 so this completes our session I hope the solution is clear to you bye and have a nice day