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Calculating derivatives is a crucial skill in maximizing present value, particularly because it helps determine the rate of change of the present value with respect to time. In this context, the present value of the car collection is given by a function, and we utilize this function to establish how the value changes as time goes by. The derivative, denoted as \( p'(t) \), is determined using the product rule, which is necessary when dealing with products of functions. Let’s break it down:

• The function combines both exponential and polynomial components.
• The derivative of the polynomial \( 300,000 + 1,000t^2 \) is \( 2,000t \).
• The derivative of the exponential function \( e^{-0.05t} \) is \( -0.05e^{-0.05t} \).

Putting it all together, the derivative of the present value is given by:
\[ p'(t) = (2,000t) e^{-0.05t} - 0.05 (300,000 + 1,000t^2)e^{-0.05t} \].
This equation helps us understand how the present value changes over time.

The concept of discounted cash flow is vital in determining present values by considering the time value of money.
The essence of this calculation is to adjust future cash flows, taking into account a continuous rate of inflation or discount rate, which is 5% in our problem.Understand that:

• The future value would typically increase over time due to asset appreciation.
• However, this future value needs to be adjusted to reflect its worth in present terms.
• The formula \( p = v e^{-0.05t} \) accounts for this by discounting the future value \( v \) back to the present.

This adjustment accounts for money's changing value over time, thus allowing for a more accurate comparison of different potential cash flows. Discounted cash flow is extensively used in finance to evaluate investment opportunities, allowing investors to determine the value of future earnings in today’s terms.

Identifying critical points is essential in optimization problems, such as determining when the dealership should sell its vehicles. A critical point occurs where the derivative, \( p'(t) \), is zero or undefined, indicating potential maxima or minima in the function.For our scenario:

• We set the derivative \( (2000t - 0.05(300000 + 1000t^2))e^{-0.05t} = 0 \).
• Since the exponential function \( e^{-0.05t} \) is never zero, we focus on the polynomial part.
• This reduces the problem to solving the equation \( 2000t - 0.05(300000 + 1000t^2) = 0 \).

The solution to this equation within our constraints (\( t\geq5 \)) gives us the critical point. Here, this calculation reveals \( t \approx 7.06 \), suggesting that 7.06 years is a potentially optimal time for selling to maximize present value.

The concavity of a function at a critical point helps us determine if this point is a maximum or minimum. This is where the second derivative test comes into play. The second derivative, \( p''(t) \), provides insight into the curvature of the present value function:\
\[ p''(t) = 10(20 - 5t)e^{-0.05t} \].
Evaluating this at \( t = 7.06 \):

• If \( p''(7.06) < 0 \), the function is concave down, indicating a local maximum.
• If \( p''(7.06) > 0 \), the function is concave up, indicating a local minimum.

In this exercise, \( p''(7.06) \approx -11.385 \times 10^{-11} \), which is less than zero, confirming that the critical point at \( t \approx 7.06 \) represents a maximum. This practical application of the second derivative test closes the analysis by validating that selling after approximately 7.06 years will yield the maximum discounted value.