The problem presented in the image is an application of maximizing revenue based on a pricing model. Here's the step-by-step solution:

Problem Description:

A car rental company charges $p$ dollars per day, where $60 \leq p \leq 150$. The number of cars rented per day is modeled by the linear function: $n(p) = 750 - 5p$

The company wants to determine the value of $p$ that maximizes its revenue. The revenue function is the product of the price per day ($p$) and the number of cars rented ($n(p)$).

Step 1: Define the Revenue Function

The revenue function, $R(p)$, is: $R(p) = p \cdot n(p)$ Substitute $n(p) = 750 - 5p$ into the equation: $R(p) = p \cdot (750 - 5p)$ Simplify: $R(p) = 750p - 5p^2$

Step 2: Maximize the Revenue Function

To find the maximum revenue, we take the derivative of $R(p)$ with respect to $p$ and set it equal to zero: $\frac{dR}{dp} = 750 - 10p$ Set $\frac{dR}{dp} = 0$: $750 - 10p = 0$ Solve for $p$: $p = 75$

Step 3: Verify the Maximum

The second derivative of $R(p)$ is: $\frac{d^2R}{dp^2} = -10$ Since the second derivative is negative ($-10 < 0$), the revenue function $R(p)$ has a maximum at $p = 75$.

Step 4: Conclusion

The company should charge $75 per day to maximize its revenue.

Let me know if you'd like a more detailed explanation or have additional questions! Here are some related questions for further exploration:

1. What is the maximum revenue the company can earn?
2. How many cars are rented when the price is $75?
3. What happens to the revenue if the price is increased to $100?
4. How would changes in the $n(p)$ function affect the optimal price?
5. Can the model be applied to different price ranges (e.g., $p \leq 60$ or $p \geq 150$)?

Tip: Always verify the range of the variable (in this case, $p$) when solving maximization problems.