Polynomials

Question

A Charter Flight charges a fare of $300 per person plus $6 per person for each unsold seat on a plane. If the plane holds 100 passengers and if X represents the number of unsold seats, find the following:

 

A.) The expression for the total revnue received on the flight

B.) # of unsold seats that will produce the maximum revenue

C.) The maximum revenue

Answer 1

Let's let p = # people on the flight

x = # empty seats

r = total revenue

 

x + p = 100 because ther are 100 total seats for passengers.

 

For revenue we have 300p + 6xp

because we make $300 per passenger plus $6 per empty seat (x) per passenger (p)....   So we have two equations..

 

x+p=100                 equation 1

300p + 6xp = r        equation 2   revenue earned

 

in equation 1 we can solve for p in terms of x

then plug that value into equation 2 to give a quadratic 

showing the value of r in terms of x...

 

p = 100 - x

 

300(100-x) + 6x(100-x) = r

30000-300x+600x-6x² = r

-6x² + 300x + 30000 = r

 

This quadratic is the equation for revenue in terms of

unsold seats

 

(B).  Find # unsold seats that gives max revenue...

 

This quadratic graphs as a parabola that opens downward

because the squared term has a negative coefficient.

This means the maximum value for revenue will be the

value of r at the vertex point (x,r) of the parabola

 

To find the value of x at the vertex we can use the 

quadratic formula  ax²+bx+c   the x value at the vertex

is found by x= -b/2a.    here a=-6, b =300, c = 30000

-b/2a= -300/(-12) = 25

 

The maximum revenue is achieved when there are 25 empty seats.

 

(C) Find maximum revenue

 

We have r = -6x² + 300x  30000.  Solve forx=25

 

-6(25²)+300(25)+30000

= -6(625)+7500+30000

= 33750

 

The maximum revenue is $33,750

 

Answer 2

There are 100 seats total. Let S be the number of seats sold, and X be the number of unsold seats.

 

We know that (seats sold) + (seats unsold) = 100, or

                           S          +         X           = 100. 

 

Subtracting X from both sides, we have S = 100 - X.

 

The airline earns $300 for each seat sold. The amount it earns from seats sold is $300 * S, or 300(100 - X).

 

It also earns $6 per person present on the flight, or 6*S dollars, per unsold seat. Multiplying by the number of unsold seats gives us a total revenue, for unsold seats, of 6*S*X = 6(100-X)*X = 600*X - 6*X^2.

 

The total revenue, Y, is then

Y = (Amount earned from sold seats) + (amount earned from unsold seats)

Y = 300(100-X) + 600X - 6X^2

Y = -6X^2 + 300X + 30000

 

Since the coefficient of X^2 is negative, the parabola opens downwards, so it has a maximum (rather than a minimum) value. The maximum value (Y) and the value which maximizes it (X) are both found at the vertex. So by putting the equation in vertex form, parts B and C are solved at once.

 

Y = -6(X^2 - 50X - 5000)

Y = -6[(X-25)^2 - 625 - 5000]

Y = -6(X-25)^2 + 33750

 

The vertex is at (25,33750). When the number of unsold seats is 25, the revenue is at its maximum of $33750.