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Answer:
R(p) = -3500p^2 +48000p . . . revenue function
$6.86 . . . price for maximum revenue
Step-by-step explanation:
The 2-point form of the equation for a line can be used to find the attendance function.
y = (y2 -y1)/(x2 -x1)(x -x1) +y1
y = (27000 -20000)/(6 -8)(x -8) +20000
y = -3500(x -8) +20000
y = 48000 -3500x . . . . y seats sold at price x
The per-game revenue is the product of price and quantity sold. In functional form, this is ...
R(p) = p(48000-3500p)
R(p) = -3500p^2 +48000p . . . per game revenue
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Revenue is maximized when its derivative is zero.
R'(p) = -7000p +48000
p = 48/7 ≈ 6.86
A ticket price of $6.86 would maximize revenue.
