 Hi and how are you all today? The question says two tailors A and B earns rupees 150 and 200 per day respectively. A can stitch 6 shirts and 4 pants per day while B can stitch 10 shirts and 4 pants per day. Form a linear programming problem to minimize the labour cost to produce at least 60 shirts and 32 pants. So let us proceed with our solution. Here we have to minimize the labour cost. It means that profit is to be maximized, right? So here let Taylor A works for x days, Taylor B works for y days. So our LPP is to maximize z that is 150 x plus 200 y as A earns 150 whereas B earns 200 per day. Now here we have some constraints. So constraints that are 10 y should be greater than equal to 60. That is 6 shirts which is stitched by Taylor A and 10 shirts which are stitched by Taylor B should be greater than equal to 60. Whereas 4 pants by Taylor A and 4 pants by Taylor B should be greater than equal to 32. Then y should both be greater than equal to 0. Converting the above LPP into equations we have 6 x plus 10 y is equal to 60 which can be written as 3 x plus 5 y is equal to 30, right? And further the second one is 4 x plus 4 y is equal to 32 which can be written as x plus y is equal to 8. Now for these two equations let us find out two points that we need to plot on the graph. Now when x is 0 then the value of y is 6 whereas when y is 0 then the value of x is 10. Similarly over here we have when the value of x is 0 y is 8 and when y is 0 x is 8. Now let us plot these points on the graph. We have when x is 0 y is 6 that is 0, 6. When x is 10 y is 0. Join these two points align representing the equation that is 3 x plus 5 y is equal to 0. See on to the next equation that is this point 080. Now this line represents equation. Now these two lines are intersecting at point p where x value of x is 5 and value of y is 3. So we see that z is maximum at point 3 that we will find out by finding out the value of z at 3 points. That is at point a, p and let us say this point as b. Points a, p and b are the three points that we are dealing with. Point a has the coordinates as 0, 8. p has 5, 3 and then we have 0, sorry it was 6 and here it is 8. Now here we have the function z as 150 x plus 200. Substituting the values here we have for this point the value of z as 1200. Here we have the value of z as 1350 and here we have it as 1200. So we can clearly see that z is max coordinates are 5 and 3. Therefore, Taylor a work 5 days, Taylor b works 3 days. Right, so this ends the session. Hope you understood the whole concept of this linear programming problem. Well and have a nice day.