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 reserves at least 20 seats 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 maximise the profit for the airline. What is the maximum profit?
Let x passengers travel by executive class and y passengers travel by economy class. Let P be the profit.
Table.
|
Class |
Number of Tickets |
Profit (Rs.) |
|
Executive |
x |
1000 x |
|
Economy |
y |
600 y |
|
Total |
x + y |
1000 x + 600 y |
We are to maximise
P = 1000 x + 600 y
subject to the constrains
x + y ≤ 200
x ≥ 20
y ≤ 80
x ≥ 0, y ≥ 0
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Now we draw the graph of x + y = 200
For x = 0, y = 200
For y = 0, x = 200
∴ line meets OX in A(200, 0) and OY in L(0, 200).
x = 20 is a straight line BM parallel to y-axis at a distance of 20.
y = 80 is a straight line CN parallel to x-axis at a distance of 80.
Since feasible region satisfies all the Constraints.
∴ DEF is the feasible region.
At D(120, 80), P = 1000 × 120 + 600 × 80 = 120000 + 48000 = 168000
At E(20, 80), P = 1000 × 20 + 600 × 80 = 20000 + 48000 = 68000
At F(20, 180), P = 1000 × 20 + 600 × 180
= 20000 + 108000 = 128000
∴ maximum value = 168000 at (120, 80)
∴ maximum profit is Rs. 168000 when 120 passengers travel in executive class and 80 passengers travel in economy class.
