Problem 60 - Calculus: Early Transcendental Functions

Question

Suppose that the value of a piece of land \(t\) years from now is $$\$ 40,000 e^{2 \sqrt{r}}$$. Given \(6 \%\) annual inflation, find \(t\) that maximizes the present value of your investment: $$\$ 40,000 e^{2 \sqrt{t}-0.06 t}$$

Step-by-Step Solution

Answer

The present value of the investment is maximized at \(t = \frac{1}{0.06^2}\) years

Step 1: Define the function

Let's define the function of the present value of the investment over time: \(f(t) = 40,000 e^{2 \sqrt{t}-0.06 t}\)

Step 2: Differentiate the function

To find the maximum value we need to take derivative of the function with respect to 't': \(f'(t) = 40,000 ( e^{2 \sqrt{t}-0.06 t} ( \frac{1}{\sqrt{t}} - 0.06) ) \)

Step 3: Set derivative equal to zero

Then, in order to find the maximum, we set the derivative equal to zero and solve for 't': \(0 = 40,000 ( e^{2 \sqrt{t}-0.06 t} ( \frac{1}{\sqrt{t}} - 0.06) ) \) We drop the constant factors and solve the equation: \(0 = e^{2 \sqrt{t}-0.06 t} ( \frac{1}{\sqrt{t}} - 0.06) \)

Step 4: Solve for 't'

After simplification, and considering that the exponential term can't be zero, the equation becomes \( \frac{1}{\sqrt{t}} - 0.06 = 0 \) which implies \( t = \frac{1}{0.06^2} \)

Key Concepts

Differentiation

Differentiation is a powerful mathematical technique used to find the rate at which a function is changing at any given point. It's a basic tool in calculus that helps in finding the maxima or minima of a function. In our problem, the goal is to maximize the present value of an investment over time. Hence, differentiation is used to obtain the derivative of the present value function with respect to time, \( t \).
This is done by applying the rules of derivatives, including the chain rule, to the defined function \( f(t) = 40,000 e^{2 \sqrt{t} - 0.06 t} \). By finding the first derivative \( f'(t) \), we are essentially determining how the present value changes as time progresses. The critical points, or the values of \( t \) where \( f'(t) = 0 \), potentially indicate where the maximum present value occurs.

• Take the derivative of the function \( f(t) \) to identify the rate of change.
• Set \( f'(t) = 0 \) to find points where the rate of change is zero, signaling potential maxima or minima.
• Evaluate the points to confirm which one provides the maximum present value.

This detailed understanding and correct application of differentiation allow us to effectively find the optimal time \( t \) for investment maximization.

Exponential Function

An exponential function is one where the variable appears in the exponent, as in \( f(t) = 40,000 e^{2 \sqrt{t} - 0.06 t} \). Exponential functions are incredibly powerful in modeling growth or decay processes, such as investments, populations, and radioactive decay. In the context of the problem, the expression \( e^{2 \sqrt{t} - 0.06 t} \) reflects how the value of the land grows exponentially over time, factoring in both natural growth and inflation decay.
The key properties of exponential functions include:

• Continuous growth or decay.
• A constant rate of growth at any larger scale.

These functions are special because they are always positive, never reaching zero or negative values. In our exercise, the exponential term ensures continuity and contributes significantly to how the present value evolves over time. Moreover, understanding how changes in the exponents, such as \( 2 \sqrt{t} \) and \( -0.06 t \), affect the whole function is crucial to solving for maximum value points by calculus methods including differentiation.

Inflation Adjustment

Inflation adjustment is crucial when considering the real value of money over time. Inflation reduces the purchasing power of money, meaning that over time, the same amount of money buys fewer goods and services. In the context of investments or financial analysis, adjusting for inflation ensures that the calculated present value genuinely reflects future cash flows in today’s terms.
In the exercise, the term \( -0.06 t \) within the exponential reflects this adjustment. The 6% annual inflation rate is taken into account to correct the foresight of the investment's value in relation to time. By incorporating this term, the decrease in future value caused by inflation is part of the model.
Key considerations around inflation adjustment involve:

• Adjusting cash flow projections for expected inflation rates.
• Ensuring that investment evaluations are realistic and adjusted for time value.

This adjustment prevents potential over-valuation of the investment’s future value, and it’s why calculating the real present value gives a clearer picture of the worth of an investment after accounting for inflation.