 Hello and how are you all today? My name is Priyanka and let us discuss this question. It says an oil company requires 13,000, 20,000 and 15,000 barrels of high-grade, medium-grade and low-grade oil respectively. Refinery A produces 100, 300 and 200 barrels per day of high, medium and low-grade oil respectively whereas refinery B produces 200, 400 and 100 barrels per day respectively. If A costs Rs. 400 per day and B costs Rs. 300 per day to operate, how many days should each be run to minimise the cost requirement? Now here in this question the data which is given to us can be written in a tabular format. Now here we have the refinery we are producing high-grade, medium-grade and low-grade oil and refinery A is producing 100, 300, 200 respectively and B 200, 400, 100 respectively. Our minimum requirements are 12,000, 20,000, 15,000 and these are the costs per day that are incurred by refinery A and B. Now suppose refinery A and B should run for X and Y respectively to minimise the total cost mathematical form of the above LPP is to minimise and that is our cost function let it be 400X plus 300Y as refinery runs for X days and refinery B runs for Y days. Now the constraints that we have this is 100X plus 200Y should be greater than equal to 12,000 then 300X plus 400Y should be greater than equal to 20,000 and lastly 200X plus 100Y is greater than equal to 15,000 both X and Y should be greater than 0 right. Now the feasible region for the LPP has to be shaded by making use of these constraints. Now with the help of these constraints first of all we will be equating these constraints and find out the values for each and every equation. So for the first equation that is on dividing this whole equation by 100 we have it as X plus 2Y equal to 120 we have the points as take X as 0 then the value of Y is 60 and when we take the value of Y as 0 the value of X is 120. Similarly for the second one we have now the equation as 3X plus 4Y is equal to 200. Now if we take the value of X as 0 the value of Y comes out to be 50 and when the value of Y is 0 the value of X is 200 by 3 and lastly for the third one we have now the equation as 2X plus Y is equal to 150 when the value of X is 0 the value of Y comes out to be 150 and the value of Y is 0 the value of X is 75. Now we need to plot all these points on the graph and obtain three lines representing these three equations. So all these three lines are represented these three equations are represented here on a graph. Now the feasible region of the above LPP is represented by the shaded region in this figure. Now here the corner points of this feasible region are A2 that is this point is A2 that is 120. Then we have P this is a point P who is having a coordinates as 60, 30 B3 this is a point B3 that is having coordinates 0, 150. The value of the objective function at these points are given in the table. Now here at this point we have the value of Z as on substituting the value of X as 120 and Y as 0 we have it as 48,000 here the value of Z is coming out to be 33,000 and here it is 45,000. So we can clearly see that Z is minimum when X is 60 and Y is 30. Hence the machine when X is 60 and Y is equal to 30. Hence the machine A that is refinery A should run for 60 days, refinery B should run for 30 days to minimize the cost that will satisfy all the constraints. So hope you understood this graph well do plot your equations very neatly and have a very nice day ahead.