91Ó°ÊÓ

Congratulations, you just won the lottery! In one option presented to you, you will be paid one million dollars a year for the next 25 years. You can deposit this money in an account that will earn \(5 \%\) each year. a. Set up a differential equation that describes the rate of change in the amount of money in the account. Two factors cause the amount to grow-first, you are depositing one millon dollars per year and second, you are earning \(5 \%\) interest. b. If there is no amount of money in the account when you open it, how much money will you have in the account after 25 years? c. The second option presented to you is to take a lump sum of 10 million dollars, which you will deposit into a similar account. How much money will you have in that account after 25 years? d. Do you prefer the first or second option? Explain your thinking. e. At what time does the amount of money in the account under the first option overtake the amount of money in the account under the second option?

Step by step solution

01

Define variables and constants

Let the amount of money in the account be denoted by \( M(t) \), where \( t \) is the time in years. The annual deposit rate is \( 1,000,000 \) dollars, and the interest rate is \( 5\% \) or \(0.05\).

02

Set up the differential equation

The rate of change of the amount of money in the account \( \frac{dM}{dt} \) is influenced by two factors: the annual deposit and the interest earned. This can be modeled by the differential equation: \[ \frac{dM}{dt} = 1,000,000 + 0.05M \]

03

Solve differential equation using an integrating factor

To solve \( \frac{dM}{dt} = 1,000,000 + 0.05M \), we first rearrange it to: \( \frac{dM}{dt} - 0.05M = 1,000,000 \). The integrating factor is \( e^{-0.05t} \). Multiplying both sides by the integrating factor, we get: \[ e^{-0.05t} \frac{dM}{dt} - 0.05e^{-0.05t}M = 1,000,000e^{-0.05t} \]. Simplifying, we find: \[ \frac{d}{dt}(M e^{-0.05t}) = 1,000,000e^{-0.05t} \]

04

Integrate both sides

Integrating both sides, we obtain: \[ M e^{-0.05t} = -20,000,000e^{-0.05t} + C \]. Simplifying, \( M(t) = -20,000,000 + Ce^{0.05t} \)

05

Apply initial conditions

Using the initial condition \( M(0) = 0 \), we find \( 0 = -20,000,000 + C \), so \( C = 20,000,000 \). Thus, \[ M(t) = 20,000,000(1 - e^{-0.05t}) \].

06

Find amount after 25 years

Substitute \( t = 25 \) years into the equation: \[ M(25) = 20,000,000(1 - e^{-1.25}) \].

07

Compute and simplify

Calculating the value, we find \( e^{-1.25} \approx 0.2865 \). Thus, \[ M(25) \approx 20,000,000(1 - 0.2865) \approx 20,000,000 \times 0.7135 \approx 14,270,000 \] dollars.

08

Compute lump sum option

For the second option, we deposit 10 million dollars into a similar account earning \( 5\% \) interest per year. We use the formula for compound interest \( A = P(1 + r)^t \), where \( P = 10,000,000 \), \( r = 0.05 \), and \( t = 25 \). Thus, \[ A = 10,000,000(1 + 0.05)^{25} \]

09

Calculate lump sum amount

Calculating the value, \( (1 + 0.05)^{25} \approx 3.386 \), we find: \[ A \approx 10,000,000 \times 3.386 \approx 33,860,000 \] dollars.

10

Compare and determine preference

Comparing the two options, the lump sum of 10 million dollars results in \( 33,860,000 \) dollars after 25 years, while annual payments result in only \( 14,270,000 \) dollars.

11

Overtake time calculation

To find the time \( t \) where the amount from the first option overtakes the second option, we solve \( 20,000,000(1 - e^{-0.05t}) = 10,000,000(1 + 0.05)^t \). This requires trial and error or numerical methods.

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Key Concepts

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

Differential Equations in Finance

When dealing with money growth over time in financial contexts, differential equations are very helpful. They enable us to describe how a quantity, like the amount of money in an account, changes as a function of time. The equation set up to describe this needs to consider all factors that influence the growth. For instance, in this exercise, we have two major factors: deposits of one million dollars per year and interest earned at a rate of 5% per year.

So, the differential equation used here is: \(\frac{dM}{dt} = 1,000,000 + 0.05M\). This shows that the rate of change of money, \(\frac{dM}{dt}\), depends on both the fixed deposits and the interest on the existing money in the account. Such equations can turn into practical tools for predicting future money amounts under varying conditions.

Compound Interest

Compound interest is an essential concept in finance. It refers to earning interest on both the initial principal and the accumulated interest from earlier periods. This exercise uses the formula for compound interest: \(A = P(1 + r)^t\), where:

• \(A\) represents the amount of money accumulated over time, including interest.
• \(P\) is the principal amount (initial deposit).
• \(r\) is the annual interest rate.
• \(t\) is the time the money is invested for, in years.

In the second scenario of the exercise, we start with a lump sum of 10 million dollars, and it earns 5% interest annually over 25 years. By plugging in the values, we calculate the final amount in the account. This calculation helps to compare which investment option could be more beneficial in the long run.

Initial Conditions

Initial conditions are crucial in solving differential equations because they provide necessary starting points for calculations. An initial condition specifies the value of the function at a particular time. In this exercise, we use \(M(0) = 0\) to denote that at time \( t = 0 \), the account has no money. This information is used to find the constant of integration in the solution of the differential equation.

Once we have: \(M(t) = -20,000,000 + Ce^{0.05t}\) and apply the initial condition, we can solve for \(C\). Here, substituting \(t = 0\), we get: \(0 = -20,000,000 + C\), so \(C = 20,000,000\). Now the function describing the amount of money in the account over time is properly adjusted, and we write it as: \(M(t) = 20,000,000(1 - e^{-0.05t})\). This adjusted function can then predict the account balance at any point in time given initial conditions.

Integration Factor

An integration factor is a technique used to solve linear first-order differential equations of the form: \(\frac{dM}{dt} + P(t)M = Q(t)\). This method simplifies the equation and makes it easier to integrate. To use an integration factor, \( \mu(t) \), we compute it as \(\mu(t) = e^{\int P(t) dt}\).

In this exercise, the differential equation \(\frac{dM}{dt} - 0.05M = 1,000,000\) has \(P(t) = -0.05\). So, the integration factor is: \( \mu(t) = e^{\int -0.05 dt} = e^{-0.05t}\). By multiplying both sides of the equation by the integration factor:
\[e^{-0.05t} \frac{dM}{dt} - 0.05e^{-0.05t}M = 1,000,000e^{-0.05t}\].

This simplifies to: \[ \frac{d}{dt}(M e^{-0.05t}) = 1,000,000 e^{-0.05t} \]. Integration makes it easier to solve for \(M\), revealing how the amount of money in the account evolves over time.