Question

The value of a tract of timber is\(V(t)=100,000 e^{0.8 \sqrt{t}}\) where \(t\) is the time in years, with \(t=0\) corresponding to 1998 . If money earns interest continuously at \(10 \%,\) the present value of the timber at any time \(t\) is \(A(t)=V(t) e^{-0.10 t} .\) Find the year in which the timber should be harvested to maximize the present value function.

Short answer

To find the year in which the timber should be harvested to maximize the present value, calculate the derivative of the present value function \(A(t)\), set it to zero and solve for \(t\). Confirm that you have found a maximum using the second derivative test, and don't forget to check endpoints. Lastly, interpret your answer in terms of the year by adding the calculated value of \(t\) to 1998.

Step-by-step solution

Calculate the derivative of the function A(t)

The first step is to calculate the derivative of the function \(A(t)\) with respect to \(t\). This is given by \(A'(t) = V'(t) e^{-0.10t} - V(t) e^{-0.10t} \times 0.10\) where \(V'(t) = \frac{dV}{dt}\)

Set the derivative equal to zero and solve for t

The second step is to set the derivative equal to zero and solve for \(t\). That means we need to find the solution to \(A'(t) = 0\). This leads to an equation that can be solved to find the value of \(t\) at which the timber should be harvested.

Determine the maximum

After finding the values for \(t\) which satisfy \(A'(t) = 0\), we need to confirm which one (if any) represents a maximum for the present value function \(A(t)\). This can be done through the second derivative test: if \(A''(t) < 0\), then \(A(t)\) has a local maximum at \(t\). You also need to check the endpoints of the valid range for \(t\) to ensure we found the absolute maximum.

Interpret the result

Once the value of \(t\) that maximizes \(A(t)\) is found, we need to interpret this result in the context of the problem. Remember that \(t=0\) corresponds to 1998, so we need to add the calculated value of \(t\) to 1998 to find the year in which the timber should be harvested to maximize its present value.

Key concepts

Derivative Calculation

Understanding the derivative of a function is paramount in mathematics, particularly in the field of calculus. The derivative represents the rate at which a function's output changes as its input changes. Essentially, when we compute the derivative of a function, we are finding a new function that gives the slope of the tangent line to the original function's graph at any given point.

In the given problem, the derivative of the present value function, denoted as \( A'(t) \), is critical for determining when to harvest the timber for maximum value. The calculation involves the product rule of differentiation, since the present value function \( A(t) \) is the product of two functions: \( V(t) \), the value of the tract of timber, and \( e^{-0.10t} \), the discounting factor for the present value.

Calculating \( A'(t) \) involves finding \( V'(t) \), the derivative of \( V(t) \), and then applying the product rule. In more complex functions, such as those involving exponentials or other non-linear terms, careful application of differentiation rules is key. For students, visualizing the changes in function behavior and the interpretation of the derivative in practical terms, such as 'the rate of growth or decay', can often aid in comprehension.

Economic Applications of Calculus

Calculus is a powerful tool in economics, particularly in optimization problems. Economists use calculus to find the point at which a certain economic measure (like cost, revenue, or profit) is maximized or minimized. The process involves finding the derivative of the function that models the economic measure and then determining where this derivative equals zero, indicating a local extremum.

In the context of our example, we are applying calculus to determine the optimal time to harvest timber. By taking the derivative of the present value function and setting it to zero, we can find when the value of the timber, after accounting for the time value of money, is at its peak. This is directly related to real-world economic decisions businesses must make to ensure profitability.

The concept can be broadened to encompass various economic applications such as pricing strategies, maximizing production efficiency, and analyzing consumer behavior. For any economic scenario that can be modeled mathematically, calculus can likely provide a method to optimize outcomes. Understanding how to apply these mathematical concepts is invaluable for students studying economics, finance, and business.

Present Value Function

The present value function is a fundamental concept in finance. It calculates the current worth of a sum of money or a stream of cash flows that will be received at a future date. The present value adjusts the future value by a discount rate, which accounts for the time value of money - the principle that a dollar today is worth more than a dollar in the future due to its potential earning capacity.

The function itself, typically denoted as \( A(t) \), discounts future value by using an exponential decay model. The formula for present value is given by multiplying the future value by \( e^{-rt} \), where \( r \) is the continuous compounding interest rate, and \( t \) represents time. This expression is crucial because it reflects the decrease in value of future money. In the scenario of the timber tract, the calculation of present value allows the owner to decide when harvesting the timber will yield a return with the maximum present worth.

For students, grasping the present value is not just about learning a formula. It is about understanding the relationship between time, money, and investment returns. This concept is used widely in financial decision-making, including loan amortization, bond pricing, and investment appraisals. Comprehension of this concept greatly enhances the ability to evaluate different financial scenarios in a logically rigorous way.