 Hello and welcome to the session. In this session we discussed the following question that says an aeroplane of an airline can carry a maximum of 200 passengers, a profit of $400 is made on each first class ticket and a profit of $300 is made on each economy class ticket. The airline reserves at least 20 seats for first class, however at least four times as many passengers prefer to travel by economy class than by first class determine how many of each type of tickets must be sold in order to maximize the profit for the airline. What is the maximum profit? Let's start with the solution now. Now we have to find out the number of tickets of each type that must be sold in order to maximize the profit. So for this first we assume let X tickets of first class and Y tickets of economy class be sold to maximize the profit. As in the question we have that the airline reserves at least 20 seats for the first class. So means that X would be greater than equal to 20 as X is the number of tickets sold for the first class and the next condition given to us is that at least four times as many passengers prefer to travel by economy class than by the first class. So this means Y would be greater than equal to 4X and also Y would be greater than equal to 80 as we have that the airline reserves at least 20 seats for the first class and it's also given that at least four times as many passengers prefer to travel by economy class than by the first class. So from this we have Y greater than equal to 4X and Y greater than equal to 80. Now X plus Y would be the total number of tickets sold as it's given that the maximum capacity of the airplane is 200 passengers. So this means that the total number of tickets sold would be less than or equal to 200. Now profit obtained by selling tickets of first class would be equal to $400 which is the profit made on one first class ticket into the number of first class tickets which is X. So this would be equal to $400X. Next the profit obtained by selling Y tickets of economy class would be equal to the profit obtained by selling one ticket of economy class that is $300 into Y which is the number of economy class tickets sold. So this would be equal to $300 Y. So the total profit would be equal to $400X plus $300 Y. So we can say the profit function is given by Z equal to $400X plus $300Y. Also as the number of tickets cannot be negative therefore we also have that X greater than equal to zero and Y greater than equal to zero. Now as the total profit is to be maximized so the given linear programming problem can be written as maximize Z equal to $400X plus $300Y. Subject to the constraints that is these are the constraints X greater than equal to 20, Y greater than equal to 4X, Y greater than equal to 80, X plus Y less than equal to $200 then the non-negative constraints X greater than equal to zero and Y greater than equal to zero. Now first of all we will find the feasible region for this first of all we draw the graphs of X equal to 20, Y equal to 4X, Y equal to 80, X plus Y equal to $200. So we have drawn the graphs this is the line X equal to 20, this is the line Y equal to 80, this is the line X plus Y equal to $200 and this line shows the equation Y equal to 4X. Now consider the in equation X greater than equal to 20, now as you can see that the point 00 does not satisfy X greater than equal to 20 since we put X as zero we get 0 greater than equal to 20 which is not true. Therefore the origin 00 does not lie in the region X greater than equal to 20. So this line X equal to 20 and the region 2 is right hand side and above the X axis represents X greater than equal to 20. Now consider the in equation Y greater than equal to 4X, now when we put 00 in this that is X0 and Y0 we get 0 greater than equal to 0 which is true therefore we can say the origin 00 satisfies the in equation Y greater than equal to 4X. And thus the origin lies in the region Y greater than equal to 4X. So the line Y equal to 4X and the region 2 is right and above the X axis represents the in equation Y greater than equal to 4X. Next we have the in equation Y greater than equal to 80. Now we observe that the point 00 does not satisfy Y greater than equal to 80 therefore we say that the origin O does not lie in the region Y greater than equal to 80. So this line Y equal to 80 and the region above this but to the right hand side of the Y axis since Y is greater than equal to 0 would represent the in equation Y greater than equal to 80. Consider the in equation X plus Y less than equal to 200, now let's see if origin satisfies this in equation or not. Putting X equal to 0 and Y equal to 0 we have 0 plus 0 is less than equal to 200 and this is true therefore we say that the point 00 satisfies the in equation X plus Y less than equal to 4X. So this is equal to 200 and therefore the origin O lies in the region plus Y less than equal to 200. So X plus Y less than equal to 200 is represented by the line X plus Y equal to 200 and the region to its left hand side including the origin but above the X axis and to the right of the Y axis only. The region common to all these in equations is the feasible region. So this shaded region represented by the points A, B, C is the feasible region. This point A has coordinates 20, 80, point B has coordinates 40, 160 and the point C has coordinates 20, 180. So we say the shaded region A, B, C is the feasible region. Since in this problem we are supposed to maximize the profit so we would use the iso profit method. For this we will assign a constant value to the objective function Z let the objective function Z equal to 400X plus 300Y be given a constant value. Say 12,000 then we get a line 400X plus 300Y is equal to 12,000 or 4X plus 3Y is equal to 120. Now let this be equation 1. We will draw a line P1Q1 to represent this equation 1. This is the line P1Q1 which represents this equation 1. Now we will give another constant value to the objective function Z. So now let the objective function Z equal to 400X plus 300Y be given a constant value 16,000. We get a line 400X plus 300Y equal to 16,000 or 4X plus 3Y equal to 160. Let this be equation 2. Now we will draw a line P2Q2 to represent the equation 2 time P2Q2 which is parallel to P1Q1 and this represents the equation 2 which is 4X plus 3Y equal to 160. Now we will move from P1Q1 to P2Q2 and so on that is we move in the increasing direction away from the origin. So as to get a line which is farthest from the origin and has at least one point in common with the feasible region ABC. So we obtain this line P3Q3 which is parallel to the set of lines P1Q1, P2Q2 and so on. And this line has a point B common with the feasible region. Now if we move this line P3Q3 above the point B then this line will be out of the feasibility region. So this means this point B which coordinates 4160 is the farthest point away from the origin. Thus we can say that the point B which coordinates 4160 gives the maximum value of the objective function Z. Hence we can say the optimal solution equal to 40Y equal to 160. The optimal value of the objective function Z is equal to 400X that is 400 into 40 plus 300Y that is 300 into Y. And so we have here 16000 plus 48000 is the optimal value of Z which is equal to C4000. This is the optimal value of Z thus we can now say 60 tickets economy class to maximise the profit $4000. So this is our final answer. This completes the session. Hope you have understood the solution of this question.