# Find linear equation based on given costs

1. Jun 14, 2015

### ducmod

1. The problem statement, all variables and given/known data

Hello!

Please, take a look at the screeshot with a problem description.

2. Relevant equations
I am trying to solve this, but I seem to have a wrong understanding of the problem.

3. The attempt at a solution
It is said that a company charges 2.50 for the first 1/5 of a mile; therefore, depending on how many miles a passenger has ridden, it should be 2.50 x m for every mile (every first fifth of every mile ), where m = miles.

Than, it is said that every additional 1/5 costs 0.45. This means that 0.45 x m x n, where m = miles and n = number of fifths in each mile; or it can be 0.45 x n, where n = all fifth of all miles during the ride (the last mile might not be complete, thus it will not contain 4/5 as all full miles would).

The answer in the textbook is
F(m) = 2:25m+2:05 The slope 2:25 means it costs an additional $2:25 for each mile beyond the rst 0.2 miles. F(0) = 2:05, so according to the model, it would cost$2:05 for a trip of 0
miles.

How 2.25 has been computed? Why is there 2.05 (which I assume is 2.50 - 0.45)?

Thank you very much!

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2. Jun 14, 2015

### jbriggs444

The company charges 2.50 for the first 1/5 of the first mile in the trip. If the passenger is driven 1/5 of a mile, the total fee for the trip will be $2.50. That$2.50 fee is incurred only once per trip regardless of trip length.

All full miles other than the first involve fees for five fifths, not four.

The problem is not expecting you to delve into the details of what happens between each fifth of a mile -- it is not asking you to come up with a step function. Instead it asks for a linear function. I would suggest that you come up with a linear function which gives the correct fee for distances that are even multiples of 1/5 of a mile.

3. Jun 14, 2015

### HallsofIvy

Staff Emeritus
You appear to be misunderstanding the term "first fifth of a mile" apparently interpreting that to mean that, for every mile, you are charged $2.50 for the first 1/5 mile, then$0.45 for each of the next 1/5 of a mile in that mile for a total of $2.50+ 4(0.45)=$4.30 per mile. That is wrong. The "first fifth of a mile" means the first 1/5 mile of the journey. If you ride in the taxi for 4 miles, say, that would be a total of 4(6)= 24 "1/5 miles". You would pay $2.50 for the first 1/5 mile then$0.45 for each of the other 23 "1/5 miles" for a total of $2.50+ 23(0.45)=$11.75.

4. Jun 15, 2015

### ducmod

Thank you, Friends, for your help!
I got it. Though I still think that the problem is ambiguously defined.

5. Jun 15, 2015

### ducmod

I solved it this way:

0.45 x 5 = 2.25

Therefore, we have a point (1/5 ; 2.50 ) and a slope of 2.25. Solving for y, we get y = 2.25m - 2.25 x (1/5) + 2.50 = 2.25m + 2.05

6. Jun 15, 2015

### ducmod

But it is still not clear to me, why 2.05 is a constant, if 2.50 is paid for the first 1/5 of a mile, and 2.50 is constant. In terms of usual meaning, 2.05 would mean a fixed cost that has to be paid even if the ride took zero miles.

7. Jun 15, 2015

### HallsofIvy

Staff Emeritus
What? Any specific number cannot be anything but a constant! And where did you get "2.05"? I'm not clear now whether your difficulty is with mathematics or the English language.

8. Jun 15, 2015

### RUber

Your solution works for journeys longer than 1/5 of a mile. But fails for trips between 0 and 1/5 of a mile. Try plotting the cost...it is not linear. It is piecewise linear.
Maybe you should consider defining the solution piecewise:
For x in miles:
$f(x) = \left\{ \begin{array}{l l} ??? & 0 \leq x < 1/5 \\ ??? &1/5 \leq x \end{array} \right.$

9. Jun 15, 2015

### jbriggs444

How about a tabular format?

0 miles: Not applicable. No passengers will embark on or pay for such trips. Price is, accordingly, irrelevant.
1/5 mile: Pay $2.50 for first 1/5 mile for a total of$2.50.
2/5 miles: Pay $2.50 for first 1/5 mile and$0.45 for second 1/5 mile for a total of $2.95 3/5 miles: Pay$2.50 for first 1/5 mile and $0.90 for the succeeding 2/5 mile for a total of$3.40

Extrapolate to a linear function that is correct at each even multiple of 1/5 mile.