Initial Value of the Portfolio

Oct 13, 2025 by leolavanderia.com.br 67 views

Introduction

In this article, we will delve into the world of finance and mathematics to understand Jennifer's stock portfolio. The value of her portfolio is given by the function $v(t) = 50 + 73t - 3t^2$ , where $v(t)$ is the value of the portfolio in hundreds of dollars, and $t$ is the time in months. We will explore the initial value of her portfolio and the time at which it reaches a maximum value.

Initial Value of the Portfolio

To find the initial value of Jennifer's portfolio, we need to evaluate the function $v(t)$ at $t = 0$ . This is because the initial value of the portfolio is the value of the portfolio at the beginning, i.e., when $t = 0$ .

v(0) = 50 + 73(0) - 3(0)^2
v(0) = 50 + 0 - 0
v(0) = 50

Therefore, Jennifer started with a portfolio value of $50$ hundred dollars, which is equivalent to $5000$ dollars.

Maximum Value of the Portfolio

To find the time at which the portfolio reaches its maximum value, we need to find the critical points of the function $v(t)$ . The critical points are the values of $t$ at which the derivative of the function is equal to zero.

v'(t) = 73 - 6t

To find the critical points, we set the derivative equal to zero and solve for $t$ .

73 - 6t = 0
6t = 73
t = 73/6
t = 12.17

Therefore, the portfolio reaches its maximum value at $t = 12.17$ months.

Value of the Portfolio at Maximum

To find the value of the portfolio at the maximum, we substitute the value of $t$ into the function $v(t)$ .

v(12.17) = 50 + 73(12.17) - 3(12.17)^2
v(12.17) = 50 + 892.41 - 3(148.22)
v(12.17) = 50 + 892.41 - 444.66
v(12.17) = 497.75

Therefore, the maximum value of the portfolio is $497.75$ hundred dollars, which is equivalent to $49775$ dollars.

Conclusion

In this article, we have explored Jennifer's stock portfolio and found the initial value and the maximum value of the portfolio. We have also found the time at which the portfolio reaches its maximum value. The initial value of the portfolio is $50$ hundred dollars, and the maximum value is $497.75$ hundred dollars. The portfolio reaches its maximum value at $t = 12.17$ months.

Mathematical Concepts

Derivatives: The derivative of a function is a measure of how the function changes as its input changes. In this article, we used the derivative to find the critical points of the function $v(t)$ .
Critical Points: The critical points of a function are the values of the input at which the derivative is equal to zero. In this article, we used the critical points to find the time at which the portfolio reaches its maximum value.
Quadratic Functions: The function $v(t) = 50 + 73t - 3t^2$ is a quadratic function. Quadratic functions have a parabolic shape and can be written in the form $f(x) = ax^2 + bx + c$ . In this article, we used the quadratic function to model the value of Jennifer's portfolio.

Real-World Applications

Finance: The concept of derivatives and critical points can be applied to finance to model the value of stocks and bonds.
Economics: The concept of quadratic functions can be applied to economics to model the relationship between variables such as supply and demand.
Science: The concept of derivatives and critical points can be applied to science to model the behavior of physical systems such as population growth and chemical reactions.

Future Research Directions

Modeling Stock Prices: Future research can focus on modeling stock prices using quadratic functions and derivatives.
Optimizing Portfolio Value: Future research can focus on optimizing the value of a portfolio using derivatives and critical points.
Applying Mathematical Concepts to Real-World Problems: Future research can focus on applying mathematical concepts such as derivatives and quadratic functions to real-world problems in finance, economics, and science.
Q&A: Jennifer's Stock Portfolio ==============================

Frequently Asked Questions

In this article, we will answer some frequently asked questions about Jennifer's stock portfolio.

Q: What is the initial value of Jennifer's portfolio?

A: The initial value of Jennifer's portfolio is $50$ hundred dollars, which is equivalent to $5000$ dollars.

Q: When does the portfolio reach its maximum value?

A: The portfolio reaches its maximum value at $t = 12.17$ months.

Q: What is the maximum value of the portfolio?

A: The maximum value of the portfolio is $497.75$ hundred dollars, which is equivalent to $49775$ dollars.

Q: How can I calculate the value of the portfolio at any given time?

A: To calculate the value of the portfolio at any given time, you can use the function $v(t) = 50 + 73t - 3t^2$ , where $t$ is the time in months.

Q: What is the significance of the derivative of the function $v(t)$ ?

A: The derivative of the function $v(t)$ is a measure of how the function changes as its input changes. In this article, we used the derivative to find the critical points of the function $v(t)$ .

Q: What is the relationship between the derivative and the critical points of a function?

A: The derivative of a function is equal to zero at the critical points of the function. In this article, we used this relationship to find the critical points of the function $v(t)$ .

Q: How can I apply the concept of derivatives and critical points to real-world problems?

A: You can apply the concept of derivatives and critical points to real-world problems in finance, economics, and science. For example, you can use derivatives to model the value of stocks and bonds, and critical points to find the maximum or minimum value of a function.

Q: What are some real-world applications of the concept of quadratic functions?

A: Some real-world applications of the concept of quadratic functions include modeling the relationship between variables such as supply and demand, and modeling the behavior of physical systems such as population growth and chemical reactions.

Q: How can I use the concept of quadratic functions to model real-world problems?

A: You can use the concept of quadratic functions to model real-world problems by writing a quadratic function that represents the relationship between the variables of interest. For example, you can use a quadratic function to model the relationship between the price of a product and the quantity demanded.

Q: What are some common mistakes to avoid when working with derivatives and critical points?

A: Some common mistakes to avoid when working with derivatives and critical points include:

Not checking the domain of the function before finding the derivative
Not checking the derivative before finding the critical points
Not using the correct formula for the derivative
Not using the correct formula for the critical points

Q: How can I improve my understanding of derivatives and critical points?

A: You can improve your understanding of derivatives and critical points by:

Practicing problems and exercises
Watching video tutorials and online lectures
Reading textbooks and online resources
Working with a tutor or mentor

Conclusion

In this article, we have answered some frequently asked questions about Jennifer's stock portfolio. We have also discussed the concept of derivatives and critical points, and how they can be applied to real-world problems. We hope that this article has been helpful in improving your understanding of these important mathematical concepts.