Problem 22 - Applied Calculus

Question

The value of good wine increases with age. Thus, if you are a wine dealer, you have the problem of deciding whether to sell your wine now, at a price of \(\$ P\) a bottle, or to sell it later at a higher price. Suppose you know that the amount a wine-drinker is willing to pay for a bottle of this wine \(t\) years from now is \(\$ P(1+20 \sqrt{t}) .\) Assuming continuous compounding and a prevailing interest rate of 5\% per year, when is the best time to sell your wine?

Step-by-Step Solution

Answer

Sell the wine after approximately 90.72 years.

Step 1: Understanding the Problem

You are a wine dealer who needs to decide when to sell a bottle of wine. You can sell it now for $P or sell it later for a higher price \( P(1+20 \sqrt{t}) \), depending on how long you wait \( t \). The prevailing interest rate is 5% per year, compounded continuously. You need to determine when to sell to maximize the present value of the future price of the wine.

Step 2: Setting Up the Equation for Present Value

To maximize the sale, calculate the present value of the wine's future price. The formula for present value with continuous compounding is: \( \text{Present Value} = \text{Future Value} \cdot e^{-rt} \). Here, \( r \) is the interest rate (5%, or 0.05), and \( t \) is the number of years. Substitute the given future value: \( P(1+20 \sqrt{t}) \cdot e^{-0.05t} \).

Step 3: Setting Up the Product to Differentiate

The present value to be maximized is a function of \( t \): \( PV(t) = P(1+20 \sqrt{t}) \cdot e^{-0.05t} \). To find the optimal time \( t \), take the derivative of \( PV(t) \) with respect to \( t \) and set it to zero. This will give the maximum present value.

Step 4: Taking the Derivative and Solving

Calculate the derivative of \( PV(t) \):\[ \frac{d}{dt} [P(1+20 \sqrt{t}) \cdot e^{-0.05t}] = P \left( 10t^{-0.5} - 0.05(1+20 \sqrt{t}) \right) e^{-0.05t} \]Set \( \frac{dPV}{dt} = 0 \) to find critical points.Solve:\[ 10t^{-0.5} - 0.05 - t^{-0.5} = 0 \]\[ 10t^{-0.5} = 0.05 + 1 \]\[ 10t^{-0.5} = 1.05 \]Square both sides and solve for \( t \).

Step 5: Simplifying the Solution

From \( 10t^{-0.5} = 1.05 \), rearrange to get \( t = \left(\frac{10}{1.05}\right)^2 \).Calculate \( \left(\frac{10}{1.05}\right)^2 = \left(9.5238\right)^2 \approx 90.719 \). Thus, the optimal time to sell is approximately when \( t \approx 90.72 \).

Step 6: Conclusion

After solving the derivative for zero, it is determined that waiting exactly \( 90.72 \) years gives the maximum present value of the wine when compounded at 5% interest continuously. Therefore, selling the wine after \( 90.72 \) years is most beneficial.

Key Concepts

Present Value

When evaluating financial decisions over time, the concept of present value is crucial. Present value represents the current worth of a future sum of money, considering a specific interest rate. It helps assess whether a future investment or cash flow is worth as much in today's terms.
In our exercise, we want to find when selling wine maximizes its value today, rather than in the future. The formula for present value with continuous compounding is:

• \( \text{Present Value} = \text{Future Value} \cdot e^{-rt} \)

Where:

• \( e \) is the base of the natural logarithm, approximately equal to 2.71828
• \( r \) is the interest rate
• \( t \) is the time in years

For the wine dealer's problem, future value is the price wine can be sold at after some years, and it is affected by the duration you hold it. Calculating present value helps to decide if holding the wine longer or selling now is more profitable.

Continuous Compounding

Continuous compounding is a fascinating concept in finance that considers interest accumulating constantly rather than at discrete intervals, like yearly, quarterly, or monthly.
This approach means that interest is calculated and added to the principal balance at every possible moment.The formula used in continuous compounding is:

• \( A = Pe^{rt} \)

Where:

• \( A \) is the amount of money accumulated after n years, including interest
• \( P \) is the principal amount
• \( r \) is the annual interest rate
• \( t \) is the time in years

In the case of wine, the formula modifies to find present value, helping us choose when wine should be sold for maximum profit in today's terms, considering that interest is continuously compounded. This gives us a clear advantage of understanding how market values shift with time and interest.

Derivative Calculation

In optimization problems, derivatives play a central role in finding maximum or minimum values of a function. In the wine-selling exercise, derivative calculations help identify the best time to sell by addressing when the present value of future wine prices peaks.
To achieve this, calculus triggers a process:

• Define a function representing the present value of wine over time.
• Take its derivative concerning time \( t \).

Setting this derivative to zero finds critical points, helping to locate any maximum or minimum. For the wine problem, the derivative is: \[ \frac{d}{dt} [P(1+20 \sqrt{t}) \cdot e^{-0.05t}] = P \left( 10t^{-0.5} - 0.05(1+20 \sqrt{t}) \right) e^{-0.05t} \]Solving \( \frac{dPV}{dt} = 0 \) navigates us to the optimal \( t \), ensuring we sell the bottle of wine at a time that offers the highest present worth. This tangible insight into pricing over time is invaluable for making informed decisions.