A city's transit authority serves 168,000 commuters daily when the fare is $1.70.

Question

Market research has determined that every penny decrease in the fare will result in 1,050 new riders. What fare will maximize revenue?

Answer 1

Let x = # of commuters

Let y = fare amount

 

The revenue function R(x,y) = x*y, since the transit authority makes the fare amount for each commuter.

 

We would like to interrelate x and y in order to eliminate one variable from the revenue function, and then utilize either the first derivative or the vertex to find a maximum, whichever you prefer.

 

We can do that by using the given information that a penny decrease in the fare results in an increase of 1050 commuters.  This implies that the relationship is linear, and has a slope given by rise over run, which would be  -.01/1050

 

Using the standard form for a line, we have y=-.01/1050 * x + b.  Then we can use the given data point that there are 16800 commuters at a fare of $1.70 to plug in for x and y and find b.

 

1.70 = -.01/1050 * 168000 + b

3.30 = b

 

Now we have obtained the complete equation for the line, y = -.01/1050 * x + 3.30.  We can now plug in for y in the revenue function.

 

R(x) = x * (-.01/1050 * x + 3.3)

R(x) = -.01/1050 * x^2 + 3.3x

 

Vertex at 173250 commuters.  And plugging this back into the line equation, we get the fare value of $1.65.

 

Or

 

R'(x) = -.02/1050 * x + 3.3 = 0

x = 173250, which again implies y = $1.65

 

 

Answer 2

First, you will need to make a function. The input value, f, will be the fare. The output value, r, will be the number of riders:

 

f            r

1.70    168,000

1.69    169,050

1.68    170,100

...          ...

 

What is "r" in terms of "f"? I see it as "168,000 riders if the fare is $1.70, plus 1,050 riders for every penny the fare is less than $1.70."

 

r = 168,000 + 1,050(100(1.70 - f))

 

If "f" = 1.70, then the second term becomes 0, and "r" = 168,000. The extra factor of 100, in the second term, is to account for the fact that we're working in differences of pennies, rather than dollars. 100 pennies make a dollar, so multiply the difference by 100, to get it in pennies, rather than dollars (hundredths to ones).

 

Let's simplify this function, and transform it into a standard linear equation:

 

r = 168,000 + 1,050(100(1.70 - f))

r = 168,000 + 105,000(1.70 - f)

r = 168,000 + 178,500 - 105,000f

r = 346,500 - 105,000f

r = -105,000f + 346,500

 

Now, you need to make a second function. The input value, f, will be the fare. The output value, R, will be the revenue (fare * number of riders) [f * r]:

 

 f          r            R
1.70   168,000  285,600

1.69   169,050  285,694.50

1.68   170,100  285,768

...        ...           ...

 

Note: r is dependent on f.

 

What is "R" in terms of "f"? I see it as "the fare (f) multiplied by the number of riders (r, which is -105,000f + 346,500)".

 

R = f(-105,000f + 346,500)

 

Let's simplify this function, and transform it into a standard quadratic equation:

 

R = f(-105,000f + 346,500)

R = -105,000f² + 346,500f

 

We're trying to find the fare (f) that will MAXIMIZE revenue (R). The coefficient, a, here is -105,000. Since it's negative, this is a "frowny" parabola. The coefficient, b, here is 346,500. The formula for the x (f) coordinate of the vertex (peak) of a parabola is -b / 2a:

 

f = -346,500 / 2(-105,000)

f = 1.65

 

The ideal fare is, therefore, $1.65.

 

We don't need to know what the resulting revenue will be, but if we wanted to know, we would plug our "f" into the equation to get "R":

 

R = -105,000f² + 346,500f

R = -105,000(1.65)² + 346,500(1.65)

R = $285,862.50

Answer 3

Well, you may be in a calculus class after all, although this can be solved without calculus.

 

R(x) is the total revenue which is calculated by multiplying commuters times fare. So for your case,

 

R(x) = (168000 + 1050x)(1.70 - .01x)

R(x) = 285600 +105x - 10.5x²

 

Can you do the rest?