Question

The revenue of a charter bus company depends on the number of unsold seats. If 100 seats are sold, the price is $55 per seat. Each unsold seat increases the price per seat by $1. Let X represents the number of seats that are unsold.

a) Write an expression for the number of seats that are sold.
b) Write an expression for the price per seat.
c) Write an expression for the revenue
d) Find the number of unsold seats that will produce the maximum revenue.
e) Find the maximum revenue

Answer 1

if \(x\) is the number of unsold seats, \(100-x\) is the number sold. Problem doesn't explicitly state that the bus seats 100, but it is implied.

price per seat is $55 + $1*unsold seats = $55 +$1x

revenue is price per seat * seats sold

revenue should be a parabola. find the vertex of the parabola to find the number of seats unsold for maximum revenue

find the revenue with that number of unsold seats

Answer 2

can you please elaborate further.from c downwards.please

Answer 3

No, that's the part you're supposed to do! I told you what the expression for seats sold is, and what the expression for the price per seat is, and that revenue is the product of those two expressions. Surely you can figure it out from that.

Answer 4

so u mean the revenue is=$55 + $1 * 100-x?

Answer 5

55 + 100- x?

Answer 6

Revenue = (cost per seat) * (seats sold)
cost per seat = $55 + $1x
seats sold = 100-x

What is (55+1x)(100-x)?

Answer 7

ohh ok. so 55(100-x) + 1x(100-x)
=5500-55x + 100x -x^2
5500+55x -x^2
im i on the right track?

Answer 8

You need to recheck your algebra.

Answer 9

ok right noow i have first sloved for the price per seat and i had 3080 + 56x

Answer 10

No, you do not have a correct revenue equation. 5500 + 55x - x^2 is not correct. Fix that before you try to do something else.

Answer 11

after trying to fix i got -56x^2 +2520x + 308000 for the revenue equation

Answer 12

OFCOL.

\[(55+1x)(100-x) = \]

Answer 13

i got -x^2 + 55x + 5500

Answer 14

No. write it out step by step. show me your work.

Answer 15

If you keep giving the same answer, and someone keeps saying "do it again", that's a strong hint that maybe a different answer would be better :-)

Answer 16

\[(55+1x)(100-x)\]apply distributive property
\[55(100-x)+1x(100-x)\]apply distributive property again
\[55*100 - 55*x + 1x *100 - 1x^2\]now simplify and collect like terms, carefully :-)

Answer 17

ok.i first of all distributed the numbers
55(100-X) + X(100-X)
5500 - 55X +100X -X^2
5500 + 55X -X^2
=-X^2 + 55X + 5500

Answer 18

On the other hand, if you feel that buying something for $55 should get you $55 change from $100, tell me where you are working as a cashier — I'd like to buy a whole bunch of stuff from you :-)

Answer 19

-55X + 100X = ?

Answer 20

OH MY GOODNESS ..HAHAHA

Answer 21

:-)

Answer 22

"watch out, forehead, here comes palm!" :-)

Answer 23

OK SO -X^2 + 45X +5500

Answer 24

LOL.THAT WAS AN OVERSIGHT

Answer 25

yes, now we're talking!

that's a parabola, right? do you know how to find the vertex?

Answer 26

YES.
LET ME WORK IT OUT AND TELL YPU MY ANSWE THEN,

Answer 27

Sounds good.

Answer 28

OK RIGHT NOW IM BEGINNING TO GET LOST AGAIN.
WITH THE PARABOLA
a=-1
b= 45
c= 550 right?

Answer 29

where the vertex h =-b/2a
and vertex k =c-ah^2 right?

Answer 30

i meant c=5500

Answer 31

c = 5500

so the vertex is at x = -b/2a = -45/-2 =

Answer 32

yes i already got that to be 22.5.

Answer 33

and i had k=4993.75

Answer 34

Okay, how do we sell 22.5 tickets? :-)

Answer 35

we have to go with 22 or 23, but which?

Answer 36

(or not sell 22 or 23 tickets, I guess)

Answer 37

lets go with 23

Answer 38

so the 22.5 is the answer for d?

Answer 39

well, we can't sell half a ticket, so we have to go with a number that is an integral number of tickets, right? as it turns out, the result is symmetrical — revenue is the same for 22 or 23 unsold tickets. it's just that the vertex of the parabola is slightly higher, if we could only sell half tickets :-)

Answer 40

now who knows, maybe the person who wrote the question expects the nonsensical answer of 22.5...I would make a note explaining that 22.5 is the peak of the curve, but does not represent a real-life solution.

what is the revenue at 22 or 23 unsold tickets?

Answer 41

oh ok .lets say i go with 23 ryt? too find e) do i would need to substitute the 23 in the revevue equation i had.thus the -X^2 + 45X +5500 to get the maximum revenue?

Answer 42

yes, exactly.

-(23)^2 + 45(23)+5500

and try it with 22 as well, just to convince yourself that it is true, although given that we know parabolas are symmetric about the vertex and our vertex is exactly midway between 22 and 23, it shouldn't be too surprising.

Answer 43

it looks like we only lose $0.25 for not being able to sell fractional tickets — hardly worth the trouble :-)

Answer 44

oh ok .thanks very much one more question.

Answer 45

Application of Quadratic Equations

Business The manager of a bicycle shop has found that, at a price (in dollars) of p(x)=150-x/5 per bicycle, x bicycles will be sold.

a)Find an expression for the total revenue from the sale of x bicycles. (Hint: Revenue = Demand X Price.)
b)Find the number of bicycle sales that leads to maximum revenue.
c)Find the maximum revenue.

Answer 46

what did you get for revenue?

Answer 47

i havent worked for it since i know how to go about it..:-)

Answer 48

no, for the last question!

Answer 49

using 23 i had 6006

Answer 50

i had the same for 22

Answer 51

Okay, good.

Answer 52

ok now please help me with the second question