Question 1180889: A company runs food device 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=4-ln(x),5 is less than equal to x and x is less than equal to 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 hot dogs be priced to maximize the profit per game?
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's how to determine the optimal price to maximize profit:
**1. Define the Profit Function:**
* **Revenue:** Revenue is price (p) times quantity (x): R(x) = p * x = (4 - ln(x)) * x
* **Cost:** The cost is $1 per hot dog times the quantity: C(x) = 1 * x = x
* **Profit:** Profit is revenue minus cost: P(x) = R(x) - C(x) = (4 - ln(x))x - x = 4x - x*ln(x) - x = 3x - x*ln(x)
**2. Find the Derivative of the Profit Function:**
To find the maximum profit, we need to find the critical points of the profit function by taking its derivative and setting it to zero:
P'(x) = d(3x - x*ln(x))/dx = 3 - (ln(x) + x*(1/x)) = 3 - ln(x) - 1 = 2 - ln(x)
**3. Set the Derivative Equal to Zero and Solve for x:**
P'(x) = 0
2 - ln(x) = 0
ln(x) = 2
x = e² ≈ 7.389
Since x is in thousands, this means approximately 7389 hot dogs.
**4. Verify that this is a Maximum (Second Derivative Test):**
Find the second derivative of the profit function:
P''(x) = d(2 - ln(x))/dx = -1/x
Since x is always positive in our domain (5 ≤ x ≤ 500), P''(x) is always negative. A negative second derivative indicates a maximum.
**5. Find the Optimal Price:**
Substitute the value of x back into the demand equation to find the optimal price:
p = 4 - ln(e²)
p = 4 - 2
p = 2
**6. Check the endpoints:**
Since the domain of x is restricted (5 ≤ x ≤ 500), we also need to check the profit at the endpoints:
* **x = 5:** P(5) = 3(5) - 5ln(5) ≈ 15 - 8.047 ≈ 6.953
* **x = 500:** P(500) = 3(500) - 500ln(500) ≈ 1500 - 3224 ≈ -1724
The profit at x = e² is: P(e²) = 3e² - e²ln(e²) = 3e² - 2e² = e² ≈ 7.389
**Conclusion:**
The company should price the jumbo hot dogs at $2 to maximize profit. This will result in selling approximately 7389 hot dogs.
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