Question

Maximizing profit: Suppose you own a tour bus and you book groups of 20 to 70 people for a day tour. The cost per person is $30 minus $0.25 for every ticket sold. If gas and other misc costs are $200, how many tickects should you sell to maximize your profit? Treat the number of tickets as a nonnegative real number.

Answer 1

Optimization problem

Answer 2

c(x)/x = 30 - 0.25x
C(x) = c(x) + 200
= 30x - 0.25x^2 + 200
Right? If so,
C'(x) = 30 - 0.5x
C'(x) = 0 at x = 60. Answer: 60.
C''(x) < 0 at x=60 so it is indeed a max.

Answer 3

why write c(x) over x?

Answer 4

Cost per person, right?

Answer 5

ok, not sure (you are probably correct)

Answer 6

ok if I just check c(x)=[30-.25x](x)
1 person the cost would be 29.75
2 persons the cost should be 59.50
but according to the way we set it up is is 59

Answer 7

shouldn't it be c(x)=30x-.25x ?

Answer 8

I am trying to understand the problem, process and eventually the solution, any assistant is greatly appreciated

Answer 9

Good day AccessDenied ! :)

Answer 10

I don't think so, not the way you the problem is written. But perhaps I misunderstand it.

Answer 11

my understanding is that, you have an income based off of your tickets sold (a revenue), the costs for what you need (cost), and then you just create a profit function based off of it and get that maximum

hello. :)

Answer 12

well if I did not know any calculus, I would have to say unless I booked 7 people I am operating at a loss

Answer 13

And, yes, I think the "2 persons" example you gave above it should be 59. Why do you think 59.50?

Answer 14

20 people would bring me a profit of 395

70 (my ideal number) would bring me 1882.50
just not sure how to answer problem using calculus (optimization)

Answer 15

2 times 30 = 60 but minus .25 per ticket
60-.50=59.50

Answer 16

I think that maybe ignores the "cost per person." Yours there would be linear cost (cost proportional to number of people). By the way, a correction: cost, not profit, is max at x=60. Cost is min at x=20!

Answer 17

Our domain of x: 20 <= x <= 70
Revenue: R(x)
R(x)/x = 30 - 0.25x
R(x) = 30x - 0.25x^2

Costs: C(x)
C(x) = 200

P(x) = R(x) - C(x)
= (30x - 0.25x^2) - 200

P'(x) = 30 - 0.5x

Evaluate P(x) for zero of derivative + endpoints

That's what I'm thinking

Answer 18

Everything in the problem is expressed as cost, with income presumably appearing implicitly as negative cost. In this case, there is a min profit (max cost) at x=60, but a min cost at the (lower) boundary on tickets sold (always between 20 and 70). Min cost in this range is at x=20.

Answer 19

Yes, I recall with rationals
we would take
the linear equation and place it over x to spread that 200 per person
ie the more people that went on the trip reduce the 200 paid per person
C(x) = (30 -.25x -200)/x

Answer 20

... in fact, it's true whether you use c(x) and for c(x)/x above.

Answer 21

Thanks I think I will go with AccessDenied since one of my calculus books talks about
r(x) = revenue from selling x items
c(x)= cost of producing the x items
p(x)=r(x)-c(x) profit from selling x items
if p(x)=r(x)-c(x) has a max value, it occurs at which p'(x)=0

Answer 22

What I mean is, the min cost will occur at one of the boundaries in either case. Do you see it? One way the cost is convex downward, the other the cost is linear. Max profit, or min cost, will occur at x=20 or x=70.

Answer 23

Well, yes, except that this is a *constrained* optimization.

Answer 24

The extreme values can occur at places where the deriv is zero *or* at a boundary.

Answer 25

yeah, so you have to check those boundaries too x=20 and x=70

Answer 26

Thanks you two have help me soo much !!!!

Answer 27

welcome. good luck.