 Hi, and how are you all doing today? The question says a factory owner wants to produce two types of machines. A and B. For his factory, the machine A requires an area of 1000 meters square and 12 skilled men for running it. And its daily output is 50 units. Whereas the machine B requires 1200 meters square area and 8 skilled men and its daily output is 40 units. If an area of 7600 meters square and 72 skilled men be available to operate the machines, how many machines of each type should be bought to maximize the daily output? So let's proceed with the solution. Let the Y be the number of machines of type to be installed. Right, now here we need to maximize our output let it be Z. So we need to maximize Z and that is that is by machine A X that is 50 B of machine Y that is gives us 40 of output daily. But they are subjected to some constraint. The area required by type A is 1000 meters square and by type B is 12 and it should be less than equal to 7600 which is available to us. For 12 skilled men in type A and 8 skilled men in type B and should be less than equal to 72 also X and Y both should be greater than equal to 0. That means we will be dealing with the first quadrant itself. Now let us rewrite this inequality into an equation which will be now 5 X plus 6 Y equal to 38. This is the equation corresponding to this inequality. And here an equation corresponding to this inequality is 3 X plus 2 Y is equal to 18. Right, now let us find out two points with the help of which we will be plotting the graph. Now here when X is 0 then Y is 38 by 6 and when Y is 0 then X is 38 by 5. Whereas 0 then Y is 9 and when Y is 0 then X is 6. We need to plot these points on the graph 9 representing these two equations. Points on the graph we get the required region and these are the end points of this feasible region that is O P Q and R. The respective coordinates of these points are 0 0 6 0 4 3 and 0 38 by 6. Now we will be finding out these four points and it will be maximum at the point whose net value of Z is greatest. So let us find out maximum value of Z. So for finding maximum value of Z we need to find out the value of Z point 0 first. Sorry O first it is Z is equal to 50 into 0 plus 40 into 0 giving us the value of Z as 0. Now at point P coordinates are 6 0 we have 50 into 6 plus 40 into 0 that gives us the value of Z as 300. Then at Q coordinates are 4 and 3 we have 50 into 4 plus 40 into 3 giving us the answer as rupees sorry as 320 that is our output. And lastly at R now the coordinates are 0 comma 38 by 6 Z is equal to 50 into 0 plus 40 into 38 by 6 giving us the answer as 253.3. Hence we can see that the maximum value of Z that is the maximum output is when Z is equal to 20. That is point Q whose coordinates are 4 and 3 machines type A will be equal to X that is equal to 4 and type B will be equal to Y and that is 3 should be purchased. So this completes the session hope you understood the question well and do plot your points carefully and draw the diagram neat. Have a nice day ahead.