A charter bus company advertises a trip for a group as follows.
At least ten people must sign up.
The cost for the first ten participants is $60 per person.
For each member of the traveling group in excess of 10, the price for everybody's ticket is reduced by $2.
Only 25 people fit on the bus, so at most 25 people can sign up.
How many total participants will maximize the company’s revenue?

For instance, if 10 people sign up, then total revenue is $600; if 11 people sign up, then the total renuve is $638.

Question

A charter bus company advertises a trip for a group as follows.
At least ten people must sign up.
The cost for the first ten participants is $60 per person.
For each member of the traveling group in excess of 10, the price for everybody's ticket is reduced by $2.
Only 25 people fit on the bus, so at most 25 people can sign up.
How many total participants will maximize the company’s revenue? 

For instance, if 10 people sign up, then total revenue is $600; if 11 people sign up, then the total renuve is $638.

anonymous · Answer

I am getting a linear equation y=38x+220 ... this is supposed to be a calculus question.

anonymous · Answer

$$
y=\left\{\begin{matrix}60n&\forall&1\le n\le10\
n(60-2n)&\forall&11\le n\le25
\end{matrix}\right.
$$

anonymous · Answer

correction
$$y=n(60-2(n-10))=80n-2n^2$$
maximize "y"
derivative..
$$y'=80-4n=0\implies n=20
y''=-4<0
$$
hence, a number of 20 people maximizes the revenue.
Max revenue = (80)-2(20-10)^2=$1400

anonymous · Answer

MAX REVENUE = 80*20-2(20)^2=$800

anonymous · Answer

you follow?

anonymous · Answer

thanks ! got it !

anonymous · Answer

I knew it should be a quadratic ... .