Grade 10 Quadratic word problem HELP

It says:
A bus usually transports 12000 riders per day at a ticket price of $1
It wants to raise the ticket price, but for every 10 cents increase, the # of riders decreases by 400
a) What ticket price will maximize revenue?

I really don't get this
thanks

At 12,000 customers at $1.00 the revenue =$12,000
Decrease the numbers of riders for every 10 cents you raise the price

Riders x dollars = revenue

11,600 x 1.10 = $12,760
11,200 x 1.20 = $13, 446
10,800 x 1.30 = $14,040
.....And so on....
8,000 x 2.00 = $16,000
7,600 x 2.10 = $15,960
7,200x 2.20 = $15,840

Keep decreasing the customers every time you increase the $ until you start loosing money

At what dollar price do you make the most before you start to loose money

Answer is above

Hope you get it now

Good luck

OK

Let's show this -- this way:
Revenue will equal 1 + (optimal increase) time (12000 - 4000 optimal increase) ( We made it 4000 since it is every .10 increase)


R = (1+x)*(12000 - 4000x)
R = 12000 + 12000x - 4000x - 4000x^2
R = -4000x^2 +8000x + 12000 (now let's multiply by -1)
R = 4000x^2 -8000x - 12000

Now comes the tricky part. To maximize this equation, you need to take the first derivative and make it = 0. That is the point at maximization
R' = 8000x - 8000
0 = 8000x - 8000
8000 = 8000x
1 = x

So an increase of $1 will maximize revenue.

Hope that helps.

the revenue can be expressed as
R = (1+x)(12000-4000x), when x = 0, R = 12000, when x = 0.1, the ticket prices is 1.10 and the rider number decreases by 400
when x = 3, the number of riders is 0,
when x = 1, R = 16000 which is the maximum
ticket price of 2$