Question 1180982: A company runs food service concessions for sporting events throughout the country. Their marketing research department chose a particular football stadium to test 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 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 the 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 hot dog price to maximize profit:
1. **Define the cost function:**
The cost of producing *x* thousand hot dogs is simply $1 * 1000x = 1000x$ dollars.
2. **Define the revenue function:**
Revenue is the price per hot dog (p) multiplied by the number of hot dogs sold (x thousand). We're given the demand equation p = 4 - ln(x). So, the revenue function is:
R(x) = x * p = x * (4 - ln(x)) = 4x - xln(x)
3. **Define the profit function:**
Profit is revenue minus cost:
P(x) = R(x) - C(x)
P(x) = 4x - xln(x) - 1000x
P(x) = -996x - xln(x)
4. **Find the critical points:**
To maximize profit, we need to find the critical points of the profit function by taking the derivative with respect to x and setting it equal to zero:
P'(x) = -996 - (ln(x) + x * (1/x))
P'(x) = -996 - ln(x) - 1
P'(x) = -997 - ln(x)
Set P'(x) = 0:
-997 - ln(x) = 0
ln(x) = -997
x = e^(-997)
5. **Consider the domain:**
The problem states that 5 ≤ x ≤ 500. The value of x we found (e^(-997)) is extremely close to zero, much smaller than 5. It is effectively zero for any practical purposes. This means the maximum profit will occur at one of the endpoints of our interval.
6. **Evaluate profit at endpoints:**
* **x = 5:**
P(5) = -996(5) - 5ln(5) = -4980 - 8.047 = -4988.05 (approximately)
p = 4 - ln(5) ≈ 2.39
* **x = 500:**
P(500) = -996(500) - 500ln(500) = -498000 - 3224.01 = -501224 (approximately)
p = 4 - ln(500) ≈ -2.22
7. **Interpret the results:**
Although mathematically, we found a critical point, it is not within the allowed domain. The profit is higher when x=5, meaning the company should produce and sell 5000 units. The corresponding price is then 2.39 dollars.
**Conclusion:**
To maximize profit, the company should sell 5000 hot dogs at a price of approximately $2.39 each.
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