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Sagot :
The profit function is P(x) = ( - x² / 5000 ) + 13.9x - 75000.
50,000 baseball game tickets were sold at $4 per ticket.
When the price is raised to $5, then 45,000 tickets were sold.
The variable and fixed costs for the ballpark owners are $0.10 and $75,000 respectively.
Let's say x is the number of tickets sold, and P is the profit.
Then,
P = ax + b
At P = 4,
4 = (50000)a + b ---------(1)
At P = 5,
5 = (45000)a + b --------(2)
Subtracting (2) from (1),
4 - 5 = (50000)a + b - (45000)a - b
- 1 = 5000(a)
a = ( - 1/5000)
So if a = ( - 1/5000),
Then,
4 = (50000)a + b
4 = (50000)( - 1 / 5000 ) + b
4 = -10 + b
b = 14
Therefore,
p(x) = ( - x /5000) + 14
Now, the profit function is:
Profit = Revenue - Costs
P(x) = R(x) - C(x)
Now, R(x) = xp(x)
R(x) = x[ ( - x/5000) + 14]
R(x) = ( - x² / 5000 ) + 14x
The fixed cost is F(x) = $75000
Hence, the costs will be:
C(x) = 75000 + (0.10)x
Therefore the profit function is:
P(x) = R(x) - C(x)
P(x) = ( - x² / 5000 ) + 14x - 75000 - (0.10)x
P(x) = ( - x² / 5000 ) + 13.9x - 75000
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The complete question is mentioned below:
When the admission price for a baseball game was $4 per ticket, 50,000 tickets were sold. When the price was raised to $5, only 45,000 tickets were sold. Assume that the demand function is linear and that the variable and fixed costs for the ballpark owners are $0.10 and $75,000 respectively.
Find the profit P as a function of x, the number of tickets sold
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