Quadratic math problem need formula?

Question

A seminar charges 600 dollars per person, it attracts 1,000 people. For every decrease of 20 dollars the seminar attacts an additional 100 people. let x represent number of 20 dollar decreases. I need a function that will give revenue generated as a function of the number of 20 dollar decreases?

Answer 1

the revenue formula is

(600 - 20x) (1000 +100x)

which when multiplied out gives

600 000 + 40 000x - 2 000 x^2

to find the maximum point the derivative needs to be found

so revenue' (x) = 0
40 000 - 4 000 x = 0
which gives x = 10 as your answer

so to get maximum revenue you would need to decrease the price per ticket by 20x10=$200

so the ticket price would need to be
$600 - $200 = $400

you would attract a projected 1000 + 100x10 people = 2000 altogether for a maximum revenue of $800 000

Answer 2

Since the no. of $20 declines is x, there would be a total decrease of $20x per person, taking per person revenue to (600-20x). Also as each decline of $20 brings an additional 100 persons, there will be 100x additional persons taking the total to 1000+100x

Total Revenue (R) is, therefore, given by

(600-20x)*(1000+100x)
=2000(30-x)*(10+x)

Answer 3

Let the starting point be 1000 attendees at 600 dol per person. Call this revenue zero R0 = 600000. This the 0th increment

R0 = (1000+0*100)*600-0*20)

The first increment gives

R1 = (1000 +1*100)*(600-1*20)

the second increment gives

R2 = R0+(1000+2*100)(600-2*20)

the xth increment would be

Rx = (1000+x*100)*(600-x*20)

Multiplying out gives the quadratic result

Rx = 600000-40000*x+2000*x^2

Answer 4

Well, the amount charged per person is given by: 600-20x

The number of people it attracts is: 1000+100x

Therefore the revenue generated = (600-20x)(1000+100x)

Answer 5

yea i need help with that **** too! quadratic math problems suck!