Optimal Number of Passengers for Maximum Railroad Revenue

[courtrigrad]

Hello all

I am not sure whether I understood this problem:

If 40 passengers hire a special car on a train, they will be charged $8.00 each. For each passenger over the 40 this fare is cut by $0.10 apiece for all passengers. (eg 50 passengers would pay $7.00 each). How many passengers will produce the greatest income for the railroad?

My solution

(18 + 0.1n)*n = I(x)

I ' (x) = 18 - 0.1n - 0.1n

= -0.2n + 18
-0.2n = -18
n = 90

Is this correct?

Thanks a lot!

 

[Galileo]

The formula for the income isn't right. For one thing, the price increases when n becomes larger. How'd you get the 18?

BTW: I is a function of n, not x.

 

[courtrigrad]

Is it (40 + n) ( 8 - .1n) the function?

 

[NateTG]

Is it (40 + n) ( 8 - .1n) the function?

If n is the number of passengers over 40, yes. (Of course, the question asks for the total number of passengers, so you'll have to adjust...)

 

[HallsofIvy]

"If 40 passengers hire a special car on a train, they will be charged $8.00 each. For each passenger over the 40 this fare is cut by $0.10 apiece for all passengers. (eg 50 passengers would pay $7.00 each). How many passengers will produce the greatest income for the railroad?"

If the number of passengers, n, is greater than 40 then the cost will be 8- 0.1(n-40)= 8- 0.1n+ 4= 12- 0.1n.

The total revenue to the railroad would be R= (12- 0.1n)n= 12n- 0.1n².

R'= 12- 0.2n which will be 0 at an "extremum". 12- 0.2n= 0 when 0.2n= 12 or n= 60.

Clearly R'> 0 when n< 60 (for example 12- 0.2(59)= 0.2) and R'< 0 when n> 60 (for example 12- 0.2(61)= -.2) so R' is decreasing so there is a maximum at n= 60 will produce the maximum total revenue.