Quadratic Functions

Question

A transit company carries about 80 000 riders per day for a fare of $1.25. To obtain more revenue, the management plans to increase the fare. It estimates that for every 5 cent increase in fare, it will lose 1000 riders. Determine the maximum revenue.

Answer 1

Let

F = $1.25 be the current fare,

f = $0.05 be one unit of fare increment,

C = 80,000 be the current number of riders,

c = 1,000 be the unit of loss for each fare increase of f,

x be the number of units f of fare increment to be determined.

Revenue R is the product of the number of riders and the fare.

So after x units f of increment the revenue will be

R = (C - cx)(F + fx)

= CF + Cfx - cFx - cfx²

R is a quadratic function with negative coefficient of x²

therefore it is concave down, therefore

it has its maximum where its derivative is 0.

Cf - cF - 2cfx = 0

x = (Cf - cF)/2cf = C/2c - F/2f

Substituting actual numbers

x = 80000/2000 - 1.25/0.10 = 27.5

Therefore maximum revenue will be obtained by increasing the fare by $0.05 x 27.5 to a total of $2.625,

resulting in 52,500 riders.

And the maximum revenue will be $137,812.5 (as opposed to the current $100,000).

Unfortunately the company cannot have a charge involving half a cent, and needs to round.

Suppose it decides to round the fare to $2.60 or $2.65 resulting in 53,000 or 52,000 riders respectively.

In either case its revenue will be 137,800.

Suppose it decides to round the fare to $2.62 or $2.63 resulting in 52,600 or 52,400 riders respectively.

In either of those two cases its revenue will be 137,812.