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The profit function is P(x) = ( - x² / 5000 ) + 11.9x - 95000.
35,000 baseball game tickets were sold at $5 per ticket.
When the price is raised to $6, then 30,000 tickets were sold.
The variable and fixed costs for the ballpark owners are $0.10 and $95,000 respectively.
Let's say x is the number of tickets sold, and P is the profit.
Then,
P = ax + b
At P = 5,
5 = (35000)a + b ---------(1)
At P = 6,
6 = (30000)a + b --------(2)
Subtracting (2) from (1),
5 - 6 = (35000)a + b - (30000)a - b
- 1 = 5000(a)
a = ( - 1/5000)
So if a = ( - 1/5000),
Then,
5 = (35000)a + b
5 = (35000)( - 1 / 5000 ) + b
5 = - 7 + b
b = 12
Therefore,
P(x) = ( - x /5000) + 12
Now, the profit function is:
Profit = Revenue - Costs
P(x) = R(x) - C(x)
Now, R(x) = xp(x)
R(x) = x[ ( - x/5000) + 12]
R(x) = ( - x² / 5000 ) + 12x
The fixed cost is F(x) = $95000
Hence, the costs will be:
C(x) = 95000 + (0.10)x
Therefore the profit function is:
P(x) = R(x) - C(x)
P(x) = ( - x² / 5000 ) + 12x - 95000 - (0.10)x
P(x) = ( - x² / 5000 ) + 11.9x - 95000
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