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Question

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 atleast 20 seats for executive class. However atleast 4 times as many passengers prefer to travel by economy class than by the executive class. Determine how many tickets of each must be sold in order to maximize the profit for the airline. What is the maximum profit?

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Solution

Let x passengers travel by executive class and y passengers travel by economy class. We construct the following table:

ClassNumber of ticketsProfit (in Rs)Executivex1000xEconomyy600yTotalx+y1000x+600ySo, our problem is to maximize Z=1000x+600y ...(i)Subject to constraints x+y200 ...(ii) x20 ..(iii)y4x0y4x ..(iv)x0, y0 ..(v)Firstly, draw the graph of the line x+y=200x0200y2000putting (0,0)in the inequality x=y200,we have0+02000200 (which is true)So, the half plane is towards the origin.Secondly, draw the graph of the line y=4xx020y080Putting (10, 0)in the inequality y4x,we have 04×10 040 (which is false)So,the half plane is towards Xaxis.Thirdly,Draw the graph of the line x=20Putting (0, 0)in the inequality x20,we have 020 (which is false)

So, the half plane is away from the origin. Since, x, y0

So, the feasible region lies in the first quadrant.

On solving the equations, we get A(20, 80), B(40, 160) and C(20, 180).

Feasible region is ABCA.

The corner points of the feasible region are A(20, 80), B(40, 160) and C(20, 180). The value of Z at these points are as follows:

Corner point1000 x+600 yA(20, 80)68000B(40, 160)136000MaximumC(20, 180)128000

Thus, the maximum value of Z is 136000 at B(40, 160).

Thus, 40 tickets of executive class and 160 tickets of economy class should be sold to maximum the profit and the maximum profit is Rs 136000.


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