 Hi and how are you all today? My name is Priyanka and I have discussed this question. It says the company manufactures two articles A and B. There are two departments through which articles are processed, first assembly and second finishing departments. The maximum capacity of the first department is 60 hours a week and that of the other department is 48 hours a week. The production of each article A requires 4 hours in assembly and 2 hours in finishing and that of each unit of B requires 2 hours in assembly and 4 hours in finishing. If the profit is Rs 6 for each unit of A and Rs 8 for each unit of B, find the number of units of A and B to be produced per week in order to have maximum profits. Now let's proceed with the solution. Now here let X units of A are produced per week, Y units of B are produced per week. Now here we need to maximize our profit function that is Rs 6 on unit A that is X plus 8 on unit B. Subject to these constraints. Now here we know that there is a maximum of 60 hours in the first department and X A unit uses 4 hours of assembly and unit B requires 2 hours of assembly. Further in the same pattern we have 48 hours in a week for the second department so it needs to be less than equal to 48 and it requires 2 hours in finishing and B requires 4 hours in finishing. Further X and Y should be greater than 0 which then equal to 0 so these are the 3 constraints to maximize this profit function. What we need to do is we will be finding out 2 points who will be satisfying these 2 equations that are inequalities right now we will convert it into equation and then find it. Here this equation can be converted this inequality can be converted into equation dividing it by 2 we have 2X plus Y is equal to 30 and this one can be written as X plus 2Y is equal to 24. Now 2 points satisfying these 2 equations are if Y is 0 then X is 15 and if X is 0 Y is 30 and for this if X is 0 then Y is 12 and if Y is 0 then X is 24. Now we need to plot these 2 equations on the graph so these 2 equations are plotted on the graph and the shaded region is the required feasible region. Now the corner points are O let us name it as O coordinates are 0 0 P whose coordinates are 15 0 Q whose coordinates are found to be 12 and 6 and R whose coordinates are 0 12 is the required region of O P Q R. Now we will be finding out profit on all these corner points. Now the value of the profit function is 6 into 0 plus 8 into 0 the answer coming out to be 0. For this is 6 into 15 plus 8 into 0 and the answer is coming out to be rupees 90. Here the value of the profit is 6 into 12 plus 8 into 6 and the answer is coming out to be rupees 120. And lastly we have 6 into 0 plus 8 into 12 the answer coming out to be rupees 96. So we can see that profit P is maximum point Q where the coordinates are 12 and 6 which is rupees 120. So number of units of the article A should be 12 and that should of P should be 6 right. So this completes the answer hope you understood it well and enjoyed it to have a nice day.