A football team charges $30 per ticket and averages 20,000 people per game. Each person spends an average of $8 on concessions. For every drop of $1 in price, the attendance rises by 800 people. What ticket price should the team charge to maximize total revenue? Calculate the TR max.

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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.

-2x + 7 = 7

subtract 7 from both sides.

-2x = 0

divide both sides by -2.

x = 0

answer: what is the question you're trying to ask?

step-by-step explanation: