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Chapter 6: Problem 62

Forestry\(The value of a tract of timber is \)V(t)=100,000 e^{0.8} \sqrt{t}\( \)V(t)=100,000 e^{0.8} / \hat{t}\(where \)t\( is the time in years, with \)t=0\( corresponding to 2010 . If money earns interest continuously at 10\)\%\( , then the present value of the timber at any time \)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

Expert verified
To find the year where the value of the timber is maximized, we first derive the function \(A(t)=V(t)e^{-0.10t}\) and equate it to 0 to find the critical points. We then apply the second derivative test to find whether these points represent maximums.

Step by step solution

01

Find the derivative of the function

Take the derivative of the given function. We'll be using a combination of the chain rule and the product rule here, as we're working with an exponential function being multiplied by another function. The derivative of \(A(t)\) is \(A'(t) = V'(t) e^{-0.10 t} + V(t) (-0.10) e^{-0.10 t}\)
02

Simplify the derivative

Next, we simplify the derivative by substituting the given \(V(t)\) and \(V'(t)\) in \(A'(t)\). But first, we need to find \(V'(t)\) which is the derivative of \(V(t)\). The derivative \(V'(t)\) will be \(0.8 \cdot 100,000 \cdot e^{0.8} /\sqrt[3]{t^2}\). With this, \(A'(t)\) simplifies to \(A'(t) = 0.8 \cdot 100,000 \cdot e^{0.8} /\sqrt[3]{t^2} \cdot e^{-0.10 t} + 100,000 \cdot e^{0.8} /\sqrt[3]{t} \cdot (-0.10) \cdot e^{-0.10 t} \)
03

Find where the derivative equals zero

To find the value of \(t\) where the function \(A(t)\) is maximum, we equate \(A'(t)\) to 0 and solve for \(t\). This will give us the critical points for the function where the function's slope is zero, which represents the maximum or minimum of the function.
04

Determine if the derivative is a maximum

Now, we need to determine if the \(t\) value obtained from Step 3 actually represents a maximum. We will use the second derivative test for this. The second derivative of the function, \(A''(t)\), is calculated and evaluated at the critical point from Step 3. If \(A''(t)<0\), then the function \(A(t)\) has a maximum at the point \(t\). We may also find second derivative numerically difficult hence we can also use first derivative test to confirm the maximum.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Derivative
The concept of a derivative is central to calculus and optimization problems. A derivative represents the rate at which a function changes as its input changes. Think of it like a speedometer in a car鈥攊t tells you how fast something is moving. In this problem, we find the derivative of a function that models timber value over time to locate maximum or minimum values, known as critical points.

To find the derivative, particularly when multiple functions are involved, we use basic rules like the chain rule and product rule. This ensures we understand how each part of the function contributes to the overall rate of change.

Key points about derivatives include:
  • They help find where a function increases or decreases.
  • They allow you to find maximums and minimums by setting the derivative to zero.
  • They show us how quickly changes happen at any point.
Exploring the Exponential Function
Exponential functions are a unique type of function where the variable appears in the exponent. They're defined generally as \(e^x\), with \(e\) being approximately 2.718, known as Euler's number. Exponential functions model growth or decay, like population growth or radioactive decay.

In this problem, the tracts of timber are modeled with an exponential function since they grow exponentially over time. Exponential functions have distinct properties:
  • They grow rapidly as the variable increases.
  • They never reach zero, which means they always have a value greater than zero.
  • They have the unique derivative characteristic: the derivative is the function itself.

This makes them particularly useful in modeling continuous growth processes, which is why they appear in financial and population models regularly.
Using the Chain Rule
The chain rule is an essential tool in calculus for finding the derivative of composite functions (functions within functions). It allows you to differentiate complex expressions by breaking them into simpler parts.

For example, if we have a function like \(f(g(x))\), the chain rule tells us how to find its derivative: \(f'(g(x)) \cdot g'(x)\). This means you take the derivative of the outer function and multiply it by the derivative of the inner function.

The steps in applying the chain rule include:
  • Identify the inner and outer functions in your expression.
  • Find the derivative of the outer function and leave the inner function as it is.
  • Calculate the derivative of the inner function.
  • Multiply the results together to find your derivative.

In this exercise, the chain rule is used when differentiating the exponential term regarding time, as it involves two inner functions.
Applying the Product Rule
The product rule is another vital technique used when differentiating products of two functions. When two functions, say \(u(t)\) and \(v(t)\), are multiplied, the product rule provides a clear method to find the derivative of their product: \((uv)' = u'v + uv'\).

This means, to take the derivative of a product, you:
  • Find the derivative of the first function and multiply it by the second function.
  • Do the same in reverse: multiply the first function by the derivative of the second.
  • Add both results together to complete the derivation.

In this exercise, the product rule helps us differentiate the combined function involving the exponential growth of timber value and its present value discount function. It's a straightforward yet powerful tool essential for dealing with complex expressions.

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Most popular questions from this chapter

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Newton's Law of Cooling When an object is removed from a furnace and placed in an environment with a constant temperature of \(80^{\circ} \mathrm{F},\) its core temperature is \(1500^{\circ} \mathrm{F}\) . One hour after it is removed, the core temperature is \(1120^{\circ} \mathrm{F}\) . (a) Write an equation for the core temperature \(y\) of the object \(t\) thours after it is removed from the furnace. (b) What is the core temperature of the object 6 hours after it is removed from the furnace?
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