Problem 67 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

OPTIMAL HOLDING TIME Beth owns an asset whose value \(t\) years from now will be \(V(t)=2,000 e^{\sqrt{2 t}}\) dollars. If the prevailing interest rate remains constant at \(5 \%\) per year compounded continuously, when will it be most advantageous to sell the collection and invest the proceeds?

Step-by-Step Solution

Answer

The optimal time to sell the asset is after 50 years.

Step 1: Define the Present Value Function

The value of the asset in the future is given by the function \(V(t) = 2,000 e^{\sqrt{2 t}}\)

Step 2: Determine the Discount Factor

The interest rate is 5% per year compounded continuously, represented as \(e^{0.05 t}\)

Step 3: Compute the Present Value

To find the present value of the proceeds if the asset is sold at time \(t \), multiply the future value by the discount factor: \[ PV(t) = 2,000 \cdot e^{\sqrt{2 t}} \cdot e^{-0.05 t} = 2,000 \cdot e^{\sqrt{2 t} - 0.05 t} \]

Step 4: Optimize the Present Value Function

To find the optimal selling time, identify the maximum of the present value function by differentiating \(PV(t)\). First, define \[P(t) = \sqrt{2} - 0.05 t \]and find the derivative:\[ \frac{d(PV(t))}{dt} = 2,000\cdot \frac{d}{dt}(e^{\sqrt{2 t} - 0.05 t}) = 0 \]

Step 5: Solve the First Derivative Equation

Differentiate and set the equation to zero:\[\frac{d}{dt} (\sqrt{2 t} - 0.05 t) = \frac{1}{2\sqrt{2 t}} - 0.05 = 0 \]Solve for \( t \):\[ \frac{1}{2\sqrt{2 t}} = 0.05 \]\[\sqrt{2 t} = 10 \]\[2 t = 100 \]\[t = 50 \]

Step 6: Verify Solution

Verify the second derivative to ensure it's a maximum:\[\frac{d^2}{dt^2} (\sqrt{2 t} - 0.05 t) = -\frac{1}{4 t^{3/2}}\]Since this is negative, \( t = 50 \) gives a maximum for the present value function.

Key Concepts

Present Value Function

The concept of present value (PV) is crucial in financial decision-making. It allows us to determine the current worth of a sum of money that will be received or paid in the future. In our exercise, we are given the future value of Beth's asset, expressed as a function of time:

• Formula: \[ V(t) = 2,000 e^{\sqrt{2t}} \].

This equation shows how the asset's value increases over time. By using the present value function, we can find out how much this future gain is worth today. This process involves adjusting future amounts of money using the interest rate.
To start, we need to understand and define the present value function, which helps us compare future values and decide the best time to sell.

Continuous Compounding

Continuous compounding is a way to calculate interest that considers the effect of compounding every moment. The general formula for continuous compounding interest is:

• \[A = P e^{rt}\] where - \(A\) is the amount of money accumulated after time \(t\), including interest,- \(P\) is the principal amount (the initial amount),- \(e\) is the base of the natural logarithm,- \(r\) is the annual interest rate,- \(t\) is the time the money is invested for.

In our scenario, the interest rate is 5%, compounded continuously, which adjusts the future value back to present value. The corresponding discount factor is given by: \[ e^{-0.05 t} \].
Applying continuous compounding ensures a more accurate and higher final amount than with simple or regular compounding.
This continuous nature of compounding is critical in finding the most advantageous time to sell Beth's asset.

Maximum Value

Finding the maximum value involves identifying the highest point of a function. This is often done using calculus, specifically by locating the critical points through differentiation. To discover the optimal holding time for Beth's asset, we need to maximize the present value function. The steps are:

• Define the present value function: \[PV(t) = 2,000 e^{\sqrt{2t} - 0.05t}\]
• Find the derivative of the PV function and set it to 0 to find the critical points: \[ \frac{d}{dt} PV(t) = 2,000 \cdot \frac{d}{dt}( e^{\sqrt{2t} - 0.05t}) = 0 \]
• Solve for \(t\) to get potential maximum values: \[ \frac{1}{2\sqrt{2t}} - 0.05 = 0 \] results in \( t = 50 \) years.
• Finally, verify the second derivative at this point to ensure it's a maximum: \[ \frac{d^2}{dt^2}(\sqrt{2t} - 0.05t) = -\frac{1}{4t^{3/2}} \]

Since the second derivative is negative, this confirms that \(t = 50\) years gives Beth the maximum present value.
This method ensures Beth makes the most profitable decision regarding when to sell her asset.