 Hello and welcome to the session. In this session we discussed the following question which says a factory owner purchases two types of machines A and B for his factory. The requirements and the limitations for the machines are as follows. So according to this table we have that for machine A the area occupied is 1000m2, labour force required is 12 men and daily output is 60 units. For machine B, area occupied is 1200m2, labour force is 8 men and daily output is 40 units. He has maximum area of 9000m2 available and 72 skilled labourers who can operate both the machines. How many machines of each type should he buy to maximize the daily output? First let's recall what is the feasible region. The common region determined by all the constraints including the non-negative constraints greater than equal to 0, why greater than equal to 0 of a linear programming problem is called the feasible region for the problem. This is the key idea that we use for this question. Now let's move on to the solution. We suppose that x is the number of machines of type A purchased and y is the number of machines of type B purchased. Now according to this table we have that the daily output by a machine of type A is 60 units. So the daily output by x machines of type A would be 60x. Similarly we have that the daily output by a machine of type B is 40 units. So for y machines of type B the daily output would be 40 into y units. So we say the total daily output for x machines of type A y machines of type B is given by 60x plus 40y units. We take let z be equal to 60x plus 40y. Now the mathematical formulation is as follows. We have that maximize z equal to 60x plus 40y since we have that z denotes the total daily output and we need to maximize the total daily output. So we have maximize z equal to 60x plus 40y. Subject to the constraints. Now let's see what are the constraints here. Now in the question we have that we have maximum area of 9000 meter square. Now let's consider this table as you can see that for machine A the area occupied is 1000 meter square. So for x machines of type A the total area occupied would be 1000x. Similarly for y machines of type B the area occupied would be 1200y meter square and the maximum area is 9000 meter square. So one of the constraints would be 1000x plus 1200y less than equal to 9000 since the maximum area available is 9000 meter square. Now let's name this equation as equation one. Now this inequality can be further written as 5x plus 6y is less than equal to 45. Let this be equation two. Now let's consider the other constraint. According to this table we have that for machine A the labour force required is 12 men and for machine B the labour force required is 8 men and in the question we have that there are 72 skilled labourers who can operate both the machines. So for x machines of type A the labour force required would be 12x and for y machines of type B the labour force required would be 8y and so we have 12x plus 8y is less than equal to 72 since skilled labourers available for both the machines is 72 men. Now this inequality could be further written as 3x plus 2y is less than equal to 18. Let this be equation three. Now then we also have x greater than equal to 0, y greater than equal to 0 these are the non-negative constraints. Now we take this as equation four. Now we graph the inequalities two, three and four. Then we also find out the feasible region. So as you can see we have drawn the graphs for the inequalities. This is the point P this is the point of intersection of the two lines 3x plus 2y equal to 18 and 5x plus 6y equal to 45. Point P has coordinates 2.25, 5.625. Now this shaded region is the feasible region that is the common region determined by all the constraints including the non-negative constraints. So we have OAPD is the feasible region that is OAPD is the feasible region for the system. Here we observe that the feasible region is bounded. Let's consider the corner points this point OA, P and D are the corner points. Let's write down the coordinates. So we have the corner points O with coordinates 00, A with coordinates 60, P with coordinates 2.25, 5.625 and D with coordinates 0.7.5. Now we have the objective function is given by Z equal to 60x plus 40y. Now we need to evaluate the objective function at each of these corner points. So at the corner point O with coordinates 00 we have objective functions there equal to 60 into x that is 16 to 0 plus 14 to y that is 14 to 0 equal to 0. Then we have the corner point A with coordinates 60. So here we have objective functions there equal to 60 into x that is 6 plus 40 into y that is 0 and this is equal to 360. The next is the point P with coordinates 2.25, 5.625. For this point the objective function Z would be equal to 60 into 2.25 plus 40 into 5.625 and this comes out to be equal to 360. Then we have the corner point D with coordinates 0, 7.5. For this the objective function Z is equal to 60 into 0 plus 40 into 7.5 and this comes out to be equal to 300. Now here as you can see in this table that the maximum value of Z is 360 and Z is the daily output that is the total daily output. So we say the maximum daily output is 360 units. Now Z is maximum at the point A with coordinates 60 or the point P with coordinates 2.25, 5.625 that is we say that Z is maximum when x equal to 6, y equal to 0 or x equal to 2.25, y equal to 5.625. Now we know that machines cannot be in decimals. So instead of x equal to 2.25 we take x equal to 2 and y equal to 5.625 we take y equal to 6. The reason machines cannot be in decimals. Now substituting x equal to 2, y equal to 6 in z equal to 60x plus 40y we get z equal to 60 into 2 plus 40 into 6 which is equal to 120 plus 240 and this is equal to 360. So thus we have Z is maximum when x equal to 6, y equal to 0 or x equal to 2, y equal to 6. Thus the final answer is the factory owner buys either 6 machines of type A since we have x equal to 6 and we know that x is the number of machines of type A then no machine of type B since y equal to 0 in this case and y is the number of machines of type B or the other cases the factory owner buys 2 machines of type A, 6 machines of type B. So this completes the session. Hope you have understood the solution for this question.