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Sagot :
9514 1404 393
Answer:
- $23.50 per ticket
- TRmax = $793,800
Step-by-step explanation:
The attendance at price p is predicted to be ...
20000 + 800(30 -p) . . . . . . 800 more people than 20,000 for each dollar below $30
The total revenue per person is predicted to be p+8, 8 more dollars than the ticket price. The total revenue is the product of attendance and revenue per person:
TR = (p +8)(20000 +800(30 -p))
TR = (p +8)(54000 -800p) = 800(p +8)(55 -p)
The factors have zeros at -8 and 55. The maximum revenue will be had when the price is the average of these values: (-8 +55)/2 = 47/2 = 23.50.
The revenue at that price is ...
TR = 800(23.50 +8)(55 -23.50) = 800(31.50^2) = 793,800
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The team should charge $23.50 to maximize total revenue. The maximum total revenue is $793,800.
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