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

Let $x$ and $y$ be the number of first class ticket holders and economy class ticket holders respectively.
$x\ge 0$
$y\ge 0$
$x+y\le 200$
$30x+20y\le 4500$
$\Rightarrow$ $3x+2y\le 450$
Let $z$ be the total profit.

Then, $z=500x+300y$
$\therefore$ the given problem reduces to maximise the objectives functions $z=500x+300y$ subject to the constraints
$x\ge 0$ ....(i)
$y\ge 0$ .....(ii)
$x+y\le 200$ .......(iii)
$3x+2y\le 450$ .....(iv)
$x\ge 0$ and $y\ge 0$ represent the closed half-planes on the next of y-axis and above x-axis respectively.
The line corresponding to $\left(iii\right)$ is $x+y=200$
$(200,0)$ and $(0,200)$ lie on $\left(iii\right)$. The origin lies in the half-plane of $\left(iii\right)$ if $0+0\le 200$. Which is true.
$\therefore$ The closed half-plane on the next of y-axis and above x-axis respectively.
$\therefore$ the closed half-plane containing the origin in graph of $\left(iii\right)$
The line corresponding to $\left(iv\right)$ is $3x+2y-\mathrm{450....}\left(vi\right)$. $(150,0)$ and $(0,225)$ lie on $\left(vi\right)$.
The origin does not lie on $\left(iv\right)$ and it lies in the half-plane of $\left(iv\right)$ if $3\left(0\right)+2\left(0\right)\le 450$, which is true. So, the closed half-plane containing the origin in the graph of $\left(iv\right)$.
The shaded bounded region in the feasible region of the given L.P.P we use cover point method to maximise $z$. The required of the feasible region are $O(0,0),A(0,200),B(75,125)$ and $C(150,0)$
At $O(0,0),z=500\left(0\right)+300\left(0\right)=0$
At $A(0,200),z=500\left(0\right)+300\left(200\right)=60000$
At $B(75,125),z=500\left(75\right)+300\left(125\right)=37500+37500=75000$
At $C(150,0),z=500\left(150\right)+300\left(0\right)=75000$
$\therefore$ Maximize value of $z=75,000$ when either $x=75,y=125$ or $x=150,y=0$
$\therefore$ profit of airline is maximum (Rs.$75000$) where either these are $75$ first class ticket holders and $125$ economy class ticket holders or there are only $150$ first class ticket holders.