 Hello and welcome to the session. I am Deepika here. Let's discuss a question which says, An aeroplane can carry a maximum of 200 passengers, a profit of rupees 1000 is made on each executive class ticket and a profit of rupees 600 is made on each economy class ticket. The airline results at least 26 for executive class. However, at least 4 times as many passengers prefer to travel by economy class than by the executive class. Determine how many tickets of each type must be sold in order to maximize the profit for the airline. What is the maximum profit? So let's start the solution. According to the given question, an aeroplane can carry a maximum of 200 passengers. Now a profit of rupees 1000 is made on each executive class ticket and a profit of rupees 600 is made on each economy class ticket. Now we have to determine how many tickets of each type must be sold in order to maximize the profit. Let the number of executive class tickets sold be x, the number of economy class tickets sold be y. Therefore, x greater than equal to 0 and y greater than equal to 0. Now again we are given an aeroplane can carry a maximum of 200 passengers since the aeroplane can carry a maximum of 200 passengers. This implies x plus y less than equal to 200. That is since the seating capacity of the aeroplane is 200. This means the number of executive class tickets sold plus the number of economy class tickets sold should be less than equal to 200. Again we are given the airline reserves at least 20 seats for executive class since the airline reserves at least 20 seats for executive class. This implies x is greater than equal to 20 because we have assumed that x is the number of executive class ticket sold. Again according to the question at least four times as many passengers prefer to travel by economy class than by the executive class. Since it is given that at least as many passengers prefer to travel economy class than by the executive class this implies y is greater than equal to 4x because y is the number of economy class ticket sold and x is the number of executive class ticket sold. So y is greater than equal to 4x because we are given that at least four times as many passengers prefer to travel by economy class than by the executive class. Now according to the question a profit of rupees 1000 is made on each executive class ticket and a profit of rupees 600 is made on each economy class ticket. So the total profit in rupees is equal to 1000x plus 600y. So let z is equal to 1000x plus 600y. So now we have the mathematical model for the given problem. So the mathematical model of the given problem is maximize z is equal to 1000x plus 600y subject to the constraints x plus y less than equal to 200. Let us give this as number one x greater than equal to 20. Let us take this as number two y greater than equal to 4x. Let us take this as number three and x greater than equal to zero and y greater than equal to zero let us take this as number four. So z is equal to 1000x plus 600y is our objective function. We have to maximize z subject to these constraints. Now we will draw the graph and find the feasible region subject to these given constraints. So first we will draw the line representing the equation x plus y is equal to 200 corresponding to the inequality x plus y less than equal to 200. Now clearly the points 0 200 and 200 0 lie on the line x plus y is equal to 200. Therefore the graph of this line can be drawn by blocking points 0 200 and 200 0 and then joining them let us take a as the point 0 200 and b as the point 200 0. So a b represents the equation x plus y is equal to 200. Now the line a b divides the plane into two half planes. So we will consider the half plane which will satisfy one. Now clearly origin satisfy this inequality. So we will consider the half plane which contains the origin. Hence the half plane which contains the origin is the graph of 1. Now again the equation corresponding to the inequality x greater than equal to 20 is x is equal to 20, x is equal to 20 represents a line parallel to y axis on the graph which passes through 20 0. Now let us take c as the point 20 0. So c d divides the plane into two half planes. So again we will consider the half plane which will satisfy two. Hence the half plane which does not contain the origin is the graph of 2. Now in a similar way we will represent y greater than equal to 4 x graphically by drawing the line y is equal to 4 x. Now clearly the points 0 0 and 20 80 satisfy the equation y is equal to 4 x. So we will plot these points on the same graph and then we will join them. Now let us take m as the point 20 80. Now the coordinates of the origin are 0 0. So again this line divides the plane into two half planes. So we will consider the half plane which will satisfy 3. So we will consider the half plane to the left of this line. Again x greater than equal to 0 and y greater than equal to 0 implies that the graph lies in the first quadrant only. Hence the shaded region in the graph is the feasible region satisfying all the given constraints. Clearly here the feasible region is triangle. Now we will find out the coordinate of its vertices. Now let us take this point as L. So clearly the coordinates of L are 20 180. Again let us take this point as n. So the coordinates of n are 40 160. The feasible region is triangle with coordinates of its vertices as whose coordinates are 2180, m whose coordinates are 20 80 and n whose coordinates are 41 60. So according to the corner point method maximum value of z will occur at any of these points. So we will calculate z which is given by 1000 x plus 600 y at each of these points. So at the point 20 180 z is equal to 1000 x 20 plus 600 x 180 and this is equal to 20000 plus 10000. Now this is equal to 10000 x 28000. Now again at the point 20 80 z is equal to 1000 x 20 plus 600 x 180 this is equal to 20000 plus 48000 this is equal to 68000. Now again at the point 41 60 z is equal to 1000 x 40 plus 600 x 160 and this is equal to 40000 plus 96000 and this is equal to 1000 x 36000. So this implies z is maximum when x is equal to 40 and y is equal to 160. Hence 40 tickets executive class 160 tickets of economy class must be sold in order to maximize the profit for the airline. Then maximum profit will be rupees 136000. So this is the answer for question. This completes our session. I hope the solution is clear to you and have a nice day.