Question

A travel agency is organising charter flight . the flight has quote a price of 300 per person if 100 or fewer sign up for flight. For every person more than 100 the price for all decrease by rs 2.50, let x be the no. of person. what value of x results for the maximum value of total ticket revenue R ?

[answer : 110]

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that

Number of passengers exceeds 100 be y.

So,

Total number of passengers in flight, x = 100 + y

Since,

The flight has quote a price of 300 per person if 100 or fewer sign up for flight. For every person more than 100 the price for all decrease by Rs 2.50.

So,

Ticket Price per passenger is = 300 - 2.5 y

So,

Total revenue received, R is given by

$\rm :\longmapsto\:R = (100 + y)(300 - 2.5y)$

$\rm :\longmapsto\:R = 30000 - 250y + 300y - {2.5y}^{2}$

$\rm :\longmapsto\:R = 30000 + 50y - {2.5y}^{2} - - - (1)$

On differentiating both sides w. r. t. y, we get

$\rm :\longmapsto\:\dfrac{d}{dy} R =\dfrac{d}{dy}( 30000 + 50y - {2.5y}^{2})$

$\rm :\longmapsto\:\dfrac{dR}{dy} =\dfrac{d}{dy} 30000 + 50\dfrac{d}{dy}y- 2.5\dfrac{d}{dy}{y}^{2}$

$\rm :\longmapsto\:\dfrac{dR}{dy} = 50 - 2.5(2y)$

$\rm :\longmapsto\:\dfrac{dR}{dy} = 50 - 5y - - - (2)$

For maxima or minima,

$\rm :\longmapsto\:\dfrac{dR}{dy} = 0$

$\rm :\longmapsto\:50 - 5y = 0$

$\rm :\longmapsto\:50 = 5y$

$\rm :\longmapsto\:y = 10$

Now,

On differentiating equation (2) w. r. t. y, we get

$\rm :\longmapsto\:\dfrac{d^{2}R}{d {y}^{2} } = - 5 < 0$

$\bf\implies \:R \: is \: maximum \: when \: y = 10$

So,

Number of passengers, x = 100 + y = 100 + 10 = 110