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Park and fly, (P& F) are a business where you pay to park your car near to the Airport and they take you to the airport and when you come back, they pick you up at the airport and you pick up your car and go home. P& F charges $8.50 per day per car but has a fixed cost of $65,000. Also, the variable cost is $1 per car and the owner pays a monthly rent for his house of $1,000.a. Find the break-even point b. Find the price that produces the maximum revenues c. Find the price that produces the maximum profit d. What is the amount of sales that P&F must make to obtain a$60,000 in profit

User Serhii Kyslyi
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1 Answer

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Cost, Revenue, and Profit

The cost function (C) usually comes in two parts: A fixed cost and a variable cost.

The revenue (or income R) is the unit price of the service times the amount of service provided.

The profit function (P) is the difference between these two:

P = R - C

The optimum operating point of the business is where the profit is positive and maximum.

The fixed cost for P&F is:

$65,000 + $1,000 = $66,000

The variable daily cost is:

$1.x per car

Where x is the number of cars it gives service to. If P&F operates 30 days a month, then the variable monthly cost is:

$30.x

The total monthly cost for P&F is:

C = $66,000 + $30x

The revenue comes from taking cars in and out of the airport for which it charges $8.50 per day per car. If x cars are served each day for a month, then the total revenue is:

R = $8.50 * 30 * x

R = $255x

a. The break-even point occurs when the cost and the revenue are equal, thus:

$255x = $66,000 + $30x

Subtracting 30x:

225x = 66,000

Dividing by 225:

x = 293.3

Thus, P&F needs to service 294 cars per day to break-even

b. Since the revenue is a linear function of the price, there is not a maximum revenue when varying the price. Revenue is higher if the price is higher at every moment. For example, if it charges $10 per car per day, the revenue would be:

R = $10 * 30 * x =$300x

c. The profit function is:

P = $255x - ($66,000 + $30x)

P = 225x - 66,000

This function is also linear with the price. The higher the price, the higher the profit. For example, charging $10 per car per day:

P = 270x - 66,000

And the break-even point will occur at x = 66,000/ 270 = 245 cars per day.

d. To obtain $60,000 in profit:

225x - 66,000 = 60,000

Adding 66,000:

225x = 66,000 + 60,000

225x = 126,000

x = 560

P&F should make 560 sales to make $60,000 in profit

User Jonasfh
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