A company runs food service concessions for sporting events throughout the country. Their marketing research department chose a particular football stadium to test market a new jumbo hot dog. It was found that the demand for the new hot dog is given approximately by
p=8−ln(x),5≤x≤500,
where x is the number of hot dogs (in thousands) that can be sold during one game at a price of p dollars. If the company pays 1 dollar for each hot dog, how should the hot dogs be priced to maximize the profit per game?
Price =
To maximize the profit per game, we need to determine the price that will maximize the profit function. The profit function can be defined as the revenue minus the cost.
Revenue = Price * Quantity
Cost = Cost per hot dog * Quantity
In this case, the Cost per hot dog is $1, and the Quantity is represented by the variable x.
Let's calculate the revenue and cost functions using the given information:
Revenue = Price * Quantity
Revenue(x) = p * x
Cost = Cost per hot dog * Quantity
Cost(x) = 1 * x
Profit = Revenue - Cost
Profit(x) = Revenue(x) - Cost(x)
Substituting the given demand function p = 8 - ln(x), we get:
Profit(x) = (8 - ln(x)) * x - 1 * x
Profit(x) = 8x - ln(x)x - x
To maximize the profit, we need to find the price that maximizes the profit function Profit(x).
In order to find the maximum value, we can take the derivative of the Profit function with respect to x, set it equal to zero, and solve for x.
Let's calculate the derivative d(Profit) / dx:
d(Profit) / dx = d(8x - ln(x)x - x) / dx
d(Profit) / dx = 8 - (1 + ln(x)) - 1
d(Profit) / dx = -ln(x) + 7
Setting the derivative equal to zero:
-ln(x) + 7 = 0
-ln(x) = -7
ln(x) = 7
Taking the exponent of both sides:
e^(ln(x)) = e^7
x = e^7
Now that we have the value of x, we can substitute it back into the demand function p = 8 - ln(x) to find the corresponding price:
p = 8 - ln(e^7)
p = 8 - 7
p = 1
Therefore, the hot dogs should be priced at $1 to maximize the profit per game.
To find the price that maximizes the profit per game, we need to determine the profit equation and then maximize it.
Given:
- Demand equation: p = 8 - ln(x), where p is the price in dollars and x is the number of hot dogs (in thousands) that can be sold during one game.
- Cost per hot dog: $1
Profit = Revenue - Cost
Revenue is calculated as the product of the price and the number of hot dogs sold:
Revenue = p * x
Cost is the number of hot dogs sold multiplied by the cost per hot dog:
Cost = $1 * x
Profit = Revenue - Cost
Profit = (p * x) - (1 * x)
Profit = (8 - ln(x)) * x - x
To find the price that maximizes the profit per game, we differentiate the profit equation with respect to x and set it equal to zero to find critical points.
d(Profit)/dx = (8 - ln(x)) - 1
Set d(Profit)/dx = 0 and solve for x:
8 - ln(x) - 1 = 0
7 - ln(x) = 0
ln(x) = 7
x = e^7
Since the range of x is limited to 5 ≤ x ≤ 500, we need to check the boundary values:
When x = 5 (lower limit), Profit = (8 - ln(5)) * 5 - 5
When x = 500 (upper limit), Profit = (8 - ln(500)) * 500 - 500
Now we compare the profits at the critical points and the boundary values to find the maximum profit.
Profit at x = e^7: (8 - ln(e^7)) * e^7 - e^7
Profit at x = 5: (8 - ln(5)) * 5 - 5
Profit at x = 500: (8 - ln(500)) * 500 - 500
The price that should be set to maximize the profit per game is the corresponding price (p) for the maximum profit value obtained from the calculations.
x = e^(8-p)
revenue is x*p = pe^(8-p)
profit = revenue-cost = xp-x = x(p-1) = (p-1)e^(8-p)
so, if y is the profit at price p,
dy/dp = (2-p)e^(8-p)
set the price at p=2