Question 1180889
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.