When the admission price for a baseball game was $5 per ticket, 35,000 tickets were sold. When the price was raised to $6, only 30,000 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 $95,000 respectively.
(a)
Find the profit P as a function of x, the number of tickets sold

Question

04-09-2022
Mathematics

Answered

When the admission price for a baseball game was $5 per ticket, 35,000 tickets were sold. When the price was raised to $6, only 30,000 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 $95,000 respectively.
(a)
Find the profit P as a function of x, the number of tickets sold

Answer :

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bharathparasad577 bharathparasad577 · Answer 1 · 2022-09-11T03:01:23-04:00

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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