Question

An aeroplane can carry a maximum load of 200 passengers. Baggage allowed to the first class ticket holder is 30 kg and for the economy class ticket holder is 20 kg. Maximum capacity of the aeroplane to carry the baggage is 4500 kg. The profit on each first class ticket is Rs.500 and on each economy class ticket is Rs.300. Formulate the problem, as L.P.P to maximize the profit.

Answer 1

Hint: For solving this problem, we should be aware about the basic concepts concerning Linear Programming problems. We will formulate the above problem in terms of two variables (x and y) to represent the above given problem. Here, x represents the number of first class ticket holders and y represents the number of economy class ticket holders. These equations are then plotted on a graph and then we get the required points where value of cost is evaluated.

Complete step-by-step answer:

Now, to solve the problem, we try to represent the constraints given in the problem using x and y, where x represents the number of first class ticket holders and y represents the number of economy class ticket holders. Thus, the first condition is that aeroplane can carry a maximum load of 200 passengers. We have,
x+y $\le $ 200 -- (1)
Now, we have the capacity for the first class ticket holder is 30 kg and for the economy class ticket holder is 20 kg. Further, the maximum capacity is given as 4500 kg. Thus, we have,
30x + 20y $\le $ 4500
3x + 2y $\le $ 450 -- (2)
Apart from these, we also know that x and y should be positive quantities. Thus, we have,
x $\ge $ 0 -- (3)
y $\ge $ 0 -- (4)
Now, to maximize profit, it is given that profit on each first class ticket is Rs.500 and on each economy class ticket is Rs.300. Thus, we have to maximize,
f(x,y) = 500x + 300y -- (A)
Now, we represent (1), (2), (3) and (4) on the graph, we have,

Now, to get the maximum profit, by finding the value of profit at the close bounded figure obtained by representing these equations on the graph. Then, we find the value of the profit at the vertices of the intersection figure formed. These vertices are (0,200), (0,0), (150,0) and intersection point of lines (1) and (2). We find the intersections of the lines (1) and (2). We get,
(solving x+y = 200 and 3x + 2y = 450) –
x = 50 and y = 150.
Now, we find profit at all these vertices. Thus, we have,
Case 1: (0,200)
f(x,y) = 500x + 300y
f(0,200) = 500 $\times $ 0 + 300 $\times $ 200
f(0,200) = 60000
Case 2: (0,0)
f(x,y) = 500x + 300y
f(0,200) = 500 $\times $ 0 + 300 $\times $ 0
f(0,200) = 0
Case 3: (150,0)
f(x,y) = 500x + 300y
f(0,200) = 500 $\times $ 150 + 300 $\times $ 0
f(0,200) = 75000
Case 4: (50,150)
f(x,y) = 500x + 300y
f(0,200) = 500 $\times $ 50 + 300 $\times $ 150
f(0,200) = 70000
Thus, the maximum profit is Rs 75000 for the case where 150 passengers are first class ticket holders and 0 passengers are economy class ticket holders.

Note: While solving LPP (Linear Programming Problems), it is always important not to forget the trivial cases (like in this problem, we had x $\ge $ 0 and y $\ge $ 0). Another, key point to note is that the graphical method can only be effectively used for solving problems concerning two variables. Suppose, we had a third constraint concerning another variable (say z), then, it becomes a 3-dimensional problem making it much more tedious to solve by graph. In such cases, generally we use a software like Matlab to solve these types of problems.