Train Fare...a mathematics question

SydneyRoverP6B

SydneyRoverP6B

Well-Known Member
Staff member
A commuter train carries 600 passengers each day from a suburb to a city. It costs $1.50 per person to ride the train. It is found that 40 fewer people will ride the train for each 5-cent increase in the fare, 40 more for each 5-cent decrease. What fare should be charged to make the largest possible revenue?

Ron.
 
rockdemon wrote,...
Hi Ron

Nice to see the return of the puzzles...

About 1.19 i make it....?
Click to expand...

Hello Rich,

Glad to ablige... :D

What did you do in order to determine your answer, as unfortunately..... :(

Ron.
 
haha :)

I just just a spreadsheet of people, fare and therefore revenue...

Only thing that could be ambiguous that i can see is that the +5% and -5% increases i've done relative to each price, not to the original 1.50 if that makes sense...

If it's a penny out either way it could be my rounding. I'll have another look later on see if i can spot my own mistake :)

Rich.
 
SydneyRoverP6B said:
A commuter train carries 600 passengers each day from a suburb to a city. It costs $1.50 per person to ride the train. It is found that 40 fewer people will ride the train for each 5-cent increase in the fare, 40 more for each 5-cent decrease. What fare should be charged to make the largest possible revenue?

Ron.
Click to expand...

If you drop the price to £1.35 you would get 680 customers and $918.00. If you dropped in further to £1.275 you would get 720 customers and again $918.00, apart from the fact that you cannot charge someone $1.275 as it is half a cent.

Richard
 
rockdemon said:
ok - I cant read. Cent. Not percent. Duh... McFly :D

I agree with Mr Quattro :)
Click to expand...

oh damn - so did I :oops:

If you drop the price to $1.15 you get 880 customers giving a total of £1,012.00, or $1.10 gives 920 customers again at $1,012.00

:?
 
Arghhh!!! More puzzles!

I also get
Commuters Fare Revenue
880 1.15 1012
920 1.1 1012

The revenue is $1,012 with 880 commuters @ $1.15, and also with 920 commuters @ $1.10!

The calculus 'method'

for every extra 0.05n cent increase, there is a 40n decrease in commuters.

Revenue (R) = (1.50 + 0.05n) x (600 - 40n)

Expanded, R = [600 x 1.50] + [600 x 0.05n] - [40n x 1.50] - [40n x 0.05n] = 900 + 30n - 60n - 2n^2 = 900 - 30n - 2n^2

R = 2 x [450 - 15n - n^2]

dR/dn = 2 x [-15 - 2n] = -2 x [15 + 2n], which is zero when 15 +2n = 0, i.e. when n = -7.5

d^2 {R}/dn^2 = -4, which is less than zero, therefore at n =-7.5, we have max value.

Of course, we cannot have non-integer in this problem (I assume!)

So, try when n = -7 and when n = -8

n = -7, 1.50 + 0.05n = 1.50 - 0.35 = 1.15, 600 - 40n = 600 + 280 = 880

R = 1.15 x 880 = 11.5 x 88 = 23 x 44 = 23 x 11 x 4 = 253 x 4 = 506 x 2 = 1012 (as found before!)

n = -8, 1.50 + 0.05n = 1.50 - 0.40 = 1.10, 600 - 40n = 600 + 320 = 920

R = 1.10 x 920 = 11.0 x 92 = (10 x 92) + 1 x 92 = 920 + 92 = 912 + 100 = 1012 (as before!)

QEV!
 
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