# An aerolpane can carry a maximum of 200 passengers. A profit of Rs. 400 is made on each first class ticket and a profit of Rs. 300 is made on each second class ticket. The airline reserves atleast 20 seats for the first class. However, at least 4 times as many passengers prefer to travel by second class, than by the first class. Determine how many tickets of each type must be sold, inorder to maximise the profit for the airline. What is the maximum profit? Make an L.P.P. and solve it graphically.

Toolbox:
• Let $R$ be the feasible region for a linear programming problem and let $Z=ax+by$ be the objective function.When $z$ has an optimum value (maximum or minimum),where variables $x$ and $y$ are subject to constraints described by linear inequalities,this optimum value must occur at a corner point of the feasible region.
• If R is bounded then the objective function Z has both a maximum and minimum value on R and each of these occur at corner points of R
Step 1:
Let the first class air tickets and second class tickets sold be $x$ and $y$
Now as the seating capacity of the aeroplane is 200,so $x+y \leq 200$
As 20 tickets for first class are to be reserved ,So we have $x\geq 20$
And as the number of tickets of second class should be at least 4 times that of first class $y \geq 4x$
Profit on sale of $x$ tickets of first class and $y$ tickets of second class $Z=1000x+600y$
Therefore LPP is (i.e) maximize $Z=1000x+600y$ subject to constraints $x+y \leq 200,x \geq 20,y\geq 4x$ and $x,y\geq 0$
Step 2:
Now let us plot the lines on the graph .
$x=y=200,x=20$ and $y=4x$
The region satisfying the inequalities $x+y\leq 200,x\geq 20$ and $y\geq 4x$ is ABC and it is shown in the figure as the shaded portion.
Step 3:
$Z=100x+600y$
The corner points of the feasible region $A(20,180),B(40,160),C(20,80)$
The values of the objective function at these points are as follows:
At the Points $(x,y)$ the value of the objective function subject to $z=1000x+600y$
At $A(20,180)$,value of the objective function $Z=1000x+600y\Rightarrow 1000\times 20+600\times 180=20000+108000=128000$
At $A(40,160)$,value of the objective function $Z=1000x+600y\Rightarrow 1000\times 40+600\times 160=40000+96000=136000$
At $A(20,80)$,value of the objective function $Z=1000x+600y\Rightarrow 1000\times 20+600\times 80=20000+48000=68000$
Step 4:
It is clear that at $B(40,160)$ $Z$ has the maximum value.
Hence $x=40,y=160$
This implies 40 tickets of first class and 160 of second class should be sold to get the maximum profit of Rs.136000.