Maximizing revenue. Edwards University wants to determine what price to charge for tickets to football games. At $18 per ticket, attendance averages 40,000 people per game. Every decrease of $3 to the ticket price adds 10,000 people to the average attendance. Every person at a game spends an average of $4.50 on concessions. What price per ticket will maximize revenue

Answer:

New price is $12.75

New attendance is 57500

Step-by-step explanation:

For starters, we use the relation.

18 - 3x

Since we don't know the number of times we're expected to deduct $3 from the price.

Again, we have 40000 + 10000x on another hand. Question says to add 10000 people everytime $3 is deducted.

Since (40000 + 10000x) is the number of people, we multiply it by $4.50, the average amount each person spends is then gotten to be

(18 - 3x) (40000 + 10000x) + (40000 + 10000x) (4.5) = 0

Expanding the bracket, we have

[720000 + 60000x - 30000x²] + [180000 + 45000x] = 0, solving further, we have

900000 + 105000x - 30000x² = 0 or

-30000x² + 105000x + 900000 = 0

Then, we find the maximum of a quadratic, by finding the axis of symmetry. Use x = -b/2a

x = -105000 / 2 * -30000

x = -105000 / -60000

x = 1.75

From our earlier equations, the new price then is

18 - 3(1.75) = 18 - 5.25 = $12.75

The new attendance also is

40000 + 10000(1.75) =

40000 + 17500 = 57500

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