• # question_answer An aeroplane can carry a maximum of 200 passengers. A profit of Rs. 1000 is made on each executive class ticket and a profit of Rs. 600 is made on each economy class ticket. The airline donate its 5% of total profit in welfare fund for poor girls. The airline reserves at least 20 seats for executive class. However, at least 4 times as many passengers prefer to travel by economy class, then by executive class. Determine how many tickets of each type must be sold in order to maximize profit for the airline? What is the maximum profit? Do you think, more passengers would prefer to travel by such an airline then by others?

Let number of executive class tickets = x             And number of economy class tickets = y Now, required linear programming problem is given by Maximise $Z=1000x+600y$ Subject to the constraints             $x+y\le 200,$ $x\ge 20,$ $y\ge 4x$ $\Rightarrow$   $4x-y\le 0\,\,\text{and}\,\,\,x,\,\,y\ge 0$ On considering the constraints as equation, we get             $x+y=200$                               ?(i)              x = 20                                     ?(ii) and       $4x-y=0$                                 ?(iii) Table for x + y = 200 is  x 0 200 y 200 0
So, line x + y = 200 passes through the points (0, 200) and (200, 0). Put (0, 0) in the inequality $x+y\le 200,$ we get $0+0\le 200$                                        [true] $\therefore$ Shaded region is towards the origin. table for$4x-y=0$  is   x 40 20 y 160 80
So, line $4x-y=0$ passes through the points (40, 160) and (20, 80).                                         Put (1, 0) in the inequality $4x-y\le 0$ we get $4(1)-0\le 0$ $\Rightarrow$ $4\le 0$                         [false] $\therefore$ Required shaded region is the region not containing (1, 0).                                          Also, the line x = 20 is parallel to Y-axis, so it passes through the point (20, 0). Put (0, 0) in the inequality $x\ge 20,$ we get $0\ge 20$                                                              [false] $\therefore$ Shaded region is away from the origin. On plotting the above points, we get the required feasible region. The intersection point of lines (i) and (ii) is (20, 180), intersection point of lines (i) and (iii) is (40, 160) and intersection point of lines (ii) and (iii) is (20, 80). Thus, corner points of the region are A(20, 180), B (40, 160) and C (20, 80). Now, consider value of Z at corner points which are given below:  Corner points $\mathbf{Z=1000x+600y}$ A(20, 180) $1000(20)+600(180)=20000$ $+108000=Rs.\text{ }128000$ B(40, 160) $1000(40)+600(160)=40000$ $+\text{ }96000=\text{ }Rs.136000\,(\text{maximum})$ C(20, 80) $1000(20)+600(80)=20000$ $+\text{ }48000=Rs.\text{ }68000$