9514 1404 393
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.
The square root of 20? You'd use Pythagorean Theorem.
Answer:
13:35 (01:35)
Step-by-step explanation:
60-13= 47
6:47
+6:48
∴ Add 48 and 47 = 95 mins
so, 95- 60= 35 mins
The extra 60 mins (1 hr) is added to the hours on the left.
so, 6+1= 7 and then 7+6 = 13
∴ 13:35 pm , which can also be written as 01:35 pm in 12-hr-clock format.