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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 ball park owners are $0.10 and $75,000 respectively.

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bharathparasad577

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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