HELP ! Algebra 2 Quadratics!?

Question

The Valley High School student council is sponsoring a dance. They anticipate that 500 students will attend. The ticket price is $3, and they estimate that for every $.20 increase in ticket price, 25 fewer students will attend. What ticket price will maximize the student council's profit?
$3.25
$3.40
$3.50
$3.80

Answer 1

I assume you mean maximize their revenue since nothing is said about the cost of the dance.

I further assume that the estimate of 500 attending is based on a ticket price of $3.00.

Let
x = number of increments of 20 cents that ticket price will be above $3.00

p = price ticket
s = number of students attending
R = revenue

p = 3 + .20x
s = 500 - 25x

R = ps

R = (3 + .20x)(500 - 25x) = 1500 - 75x + 100x - 5x²
R = -5x² + 25x + 1500
R - 1500 = -5(x² - 5x)

Complete the square.

R - 1500 - 5(5/2)² = -5[x² - 5x + (5/2)²]
R - 1531.25 = -5(x - 5/2)²

The ticket price that ensures maximum revenue is:

p = 3 + .20x = 3 + .20(5/2) = $3.50

Answer 2

The total gate receipts = (price per ticket)(number of tickets).
The price per ticket is 3 + 0.20x, where x is the number of 20¢ increases. The number of tickets sold is 500 - 25x.

g(x) = (3 + 0.20x)(500 - 25x). Multiplying,
g(x) = 1500 - 75x + 100x - 5x².
g(x) = - 5x² + 25x + 1500. To maximize the price, find the value of x for the axis of symmetry (x = -b / 2a).

x = -b / 2a
x = -(25) / [2(-5)]
x = -25 / (-10)
x = 2.5

The gate receipts are maximized when there are 2.5 of the 20¢ increases. If the tickets cost $3.50, there should be 500 - 25(2.5) = 500 - 62.5 = 437.5 students in attendance... (maybe one's a pre-schooler), but $3.50 maximizes the gate.