Question

Need Help!

A commuter railway has 800 passengers per day and charges each one two dollars per day. For each 4 cents that the fare is increased, 5 fewer people will go by train.
What is the greatest profit that can be earned?

Greatest profit = $_____

Answer 1

Explanation:

To find the greatest profit, we need to determine the fare that will maximize revenue, while also considering the decrease in ridership due to the fare increase.

Let's assume the initial fare is $2, and the number of passengers is 800 per day. So, the initial revenue is:

$2 x 800 = $1600 per day

Now, let's say we increase the fare by 4 cents to $2.04. According to the problem, for each 4 cents increase in fare, there will be 5 fewer passengers. So, the number of passengers will decrease to:

800 - (5 x 4) = 780 passengers per day

The new revenue at this fare will be:

$2.04 x 780 = $1591.20 per day

By increasing the fare, the revenue decreased. This means that we may have increased the fare too much. Let's try another fare.

If we increase the fare by 2 cents to $2.02, the number of passengers will decrease by:

800 - (5 x 2) = 790 passengers per day

The new revenue at this fare will be:

$2.02 x 790 = $1595.80 per day

This is more revenue than the initial fare of $2 per person. Let's continue this process:

If we increase the fare by another 2 cents to $2.04, the number of passengers will decrease by:

790 - (5 x 2) = 780 passengers per day

The new revenue at this fare will be:

$2.04 x 780 = $1591.20 per day

This is less revenue than the $2.02 fare, so we can stop here.

Therefore, the greatest profit can be earned by charging $2.02 per person per day, and the maximum revenue will be:

$2.02 x 790 = $1595.80 per day

This is a bit less than the initial daily revenue of $1600, but it is the most revenue we can get by increasing the fare without causing a significant reduction in ridership.

Answer 2

$2205

Explanation:

You want the greatest profit that can be earned by a commuter railway that has 800 passengers per day at a fare of $2, and 5 fewer for each 4¢ increase in the fare.

Ridership function

The number of riders (q) as a function of price (p) can be described by ...

q = 800 -5(p -2)/0.04

q = 1050 -125p . . . . . . . simplified

Revenue function

The daily revenue is the product of price and the number of riders who pay that price.

r = pq

r = p(1050 -125p)

r = 125p(8.40 -p)

Maximum revenue

This function describes a parabola that opens downward. It has zeros at p=0 and p=8.40. The vertex of the parabola is on the line of symmetry, halfway between the zeros:

pmax = (0 +8.40)/2 = 4.20

The maximum revenue is ...

r(4.20) = 125·4.20(8.40 -4.20) = 125(4.20²) = 2205

The maximum revenue that can be earned is $2205.

__

Additional comment

The ridership at that fare is 125(4.20) = 525.

Profit is the difference between revenue and cost. Here, we have no information about the cost function, so we cannot predict the maximum profit. The question seems to assume that profit is equal to revenue.