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 at least 20 seats for executive class and 80 for economy class, then the number of tickets of each class must be sold in order to maximise the profit for the airline is
$[$ where $n\left(E\right)=$ number of executive class tickets and $n\left(E^{\prime }\right)=$ number of economy class tickets $]$

• A. $n\left(E\right)=120,n\left(E^{\prime }\right)=80$
• B. $n\left(E\right)=80,n\left(E^{\prime }\right)=120$
• C. $n\left(E\right)=140,n\left(E^{\prime }\right)=60$
• D. $n\left(E\right)=60,n\left(E^{\prime }\right)=140$

Answer 1

The correct option is A $n\left(E\right)=120,n\left(E^{\prime }\right)=80$

Suppose $x$ is the number of executive class tickets and $y$ is the number of economy class tickets .
Then, total profit (in Rs) $=1000x+600y$
Let $Z=1000x+600y$
We now have the following mathematical model for the given problem.
Maximise $Z=1000x+600y\cdots \left(i\right)$
subject to the constraints:
$x+y\le 200\text{(Total seat constraint)}\cdots \left(ii\right)$
$x\ge 20\text{(executive class seat constraint)}\cdots \left(iii\right)$
$y\ge 80\text{(economy class seat constraint)}\cdots \left(iv\right)$
The feasible region (shaded) determined by the linear inequalities $\left(ii\right)\text{to}\left(iv\right)$ is shown in the Figure


Let us evaluate the objective function $Z$ at each corner point as shown below

Corner points	$Z=1000x+600y$
$(20,180)$	$128000$
$(20,80)$	$68000$
$(120,80)$	$168000$

We find that maximum value of $Z\text{is}168000\text{at}(120,80)$. Hence, $120$ tickets of executive class and $80$ tickets of economy class must be sold to realise the maximum profit and the maximum profit will be Rs $168000$.