Logistic Growth Rate Analysis

1. Extracting details: The user provides extracted text from an image, detailing a problem involving multiple parts of logistic growth.

2. Formulating steps: The user requests a solution with at least two steps, detailing given information, what needs to be found, and a step-by-step approach.

3. Presenting findings: The solution should be clearly organized, with headings for each step and a concise final answer.

Crafting the solution

I’m working through a detailed solution, breaking it into two key steps, with each step clearly defined and articulated carefully to avoid disallowed reasoning.

Understanding the problem

I’m gathering information to solve the problem. The population size is 40,000 people, and the news is spreading through word of mouth.

Tracking news diffusion

I'm examining the spread of news over time with $f(t)$. After 8 days, about 5.621 thousand people have heard it. Calculating $f'(t)$ reveals a rate of 1.0208 thousand people per day. Comparing to the logistic growth models is next.

Formulating the differential equation

I’m working out the differential equation for the news spread, expressed as $\frac{dy}{dt} = 0.4y(40 - y)$.

Determining the spread

I’m calculating the number of people reached when the news spreads at 3.99 thousand people per day, and the times at which this rate occurs.

Tracing the spread

I’m identifying the fastest rate of news spread, noting it as 160 thousand people per day, and distinguishing between two distinct values for verification.

Piecing together the scenario

I’m thinking through part (a) of the problem, which involves determining how many thousands of people have heard the news after 8 days using the given function $y = \frac{40}{1 + 150e^{-0.4t}}$.

Calculating population

OK, let's plug in t = 8 into the logistic growth equation $y = \frac{40}{1 + 150e^{-0.4 \times 8}}$. After simplifying, it turns out approximately 5.621 thousand people have heard the news after 8 days.

Mapping out the rate

I’m working through the derivative $f'(t)$ using the chain rule, aiming to simplify and determine its form. Evaluating it at $t = 6$ will pinpoint the spreading rate.

Subject: Logistic Growth and Rate of Spread Analysis

(a) Calculating the Number of People Who Have Heard the News After 8 Days

Step 1: Given Information and Objective

• Given: $y = f(t) = \dfrac{40}{1 + 150e^{-0.4t}}$
• Objective: Find $y$ when $t = 8$.

Step 2: Substitute $t = 8$ into the Equation

1. Calculate the exponent: $-0.4 \times 8 = -3.2$

2. Compute $e^{-3.2}$: $e^{-3.2} \approx 0.04076$

3. Calculate the denominator: $1 + 150 \times 0.04076 = 1 + 6.114 = 7.114$

4. Compute $y$: $y = \dfrac{40}{7.114} \approx 5.621$

Final Answer:

Approximately 5.621 thousand people have heard the news after 8 days.

(b) Calculating the Derivative $f'(t)$

Step 1: Differentiate the Given Function

• Given: $y = f(t) = \dfrac{40}{1 + 150e^{-0.4t}}$
• Objective: Find $f'(t)$.

Step 2: Apply Differentiation Techniques

1. Let $u = 1 + 150e^{-0.4t}$, so $y = \dfrac{40}{u}$.

2. Differentiate $u$ with respect to $t$: $\dfrac{du}{dt} = -60e^{-0.4t}$

3. Use the chain rule for $y$: $f'(t) = -40 \times \dfrac{du/dt}{u^2}$

4. Substitute $du/dt$ and $u$: $f'(t) = -40 \times \dfrac{-60e^{-0.4t}}{(1 + 150e^{-0.4t})^2}$

5. Simplify: $f'(t) = \dfrac{2400e^{-0.4t}}{(1 + 150e^{-0.4t})^2}$

Final Answer:

$f'(t) = \dfrac{2400e^{-0.4t}}{(1 + 150e^{-0.4t})^2}$

(c) Finding the Rate at Which the News Is Spreading After 6 Days

Step 1: Given Information and Objective

• Given: $f'(t) = \dfrac{2400e^{-0.4t}}{(1 + 150e^{-0.4t})^2}$
• Objective: Find $f'(6)$.

Step 2: Substitute $t = 6$ into $f'(t)$

1. Calculate the exponent: $-0.4 \times 6 = -2.4$

2. Compute $e^{-2.4}$: $e^{-2.4} \approx 0.09072$

3. Calculate the numerator: $2400 \times 0.09072 = 217.728$

4. Compute the denominator: $(1 + 150 \times 0.09072)^2 = (1 + 13.608)^2 = (14.608)^2 \approx 213.374$

5. Calculate $f'(6)$: $f'(6) = \dfrac{217.728}{213.374} \approx 1.0208$

Final Answer:

The news is spreading at a rate of 1.0208 thousand people per day after 6 days.

(d) Deriving the Corresponding Differential Equation

Step 1: Identify the Logistic Growth Model

• Given: $y = \dfrac{40}{1 + 150e^{-0.4t}}$
• Objective: Find the differential equation in terms of $y$.

Step 2: Recognize the Standard Logistic Equation

The standard logistic differential equation is: $\dfrac{dy}{dt} = ky(M - y)$ where:

• $k$ is the growth rate,
• $M$ is the carrying capacity.

Step 3: Match Parameters to the Given Function

• Carrying capacity $M = 40$ thousand.
• Growth rate $k = 0.4$.

Final Answer:

$\dfrac{dy}{dt} = 0.4y(40 - y)$

(e) Finding the Number of People When the Spread Rate Is 3.99 Thousand People per Day

Step 1: Set $f'(t) = 3.99$ and Solve for $y$

• Given: $f'(t) = \dfrac{2400e^{-0.4t}}{(1 + 150e^{-0.4t})^2} = 3.99$

Step 2: Relate $f'(t)$ to $y$

1. From the derivative and the logistic equation: $f'(t) = 0.4y(40 - y)$

2. Set up the equation: $0.4y(40 - y) = 3.99$

3. Simplify: $y(40 - y) = \dfrac{3.99}{0.4} = 9.975$

4. Write the quadratic equation: $y^2 - 40y + 9.975 = 0$

Step 3: Solve the Quadratic Equation

1. Use the quadratic formula: $y = \dfrac{40 \pm \sqrt{40^2 - 4 \times 1 \times 9.975}}{2}$

2. Calculate the discriminant: $\sqrt{1600 - 39.9} = \sqrt{1560.1} \approx 39.5$

3. Find the two values:

   • Smaller $y$: $y = \dfrac{40 - 39.5}{2} \approx 0.25$
   • Larger $y$: $y = \dfrac{40 + 39.5}{2} \approx 39.75$

Final Answer:

Approximately 250 people and 39,750 people have heard the news when its spread rate is 3.99 thousand people per day.

(f) Determining the Times When the Spread Rate Is 3.99 Thousand People per Day

Step 1: Use the Relationship Between $y$ and $t$

• Given: $y = \dfrac{40}{1 + 150e^{-0.4t}}$
• Objective: Find $t$ when $y = 0.25$ and $y = 39.75$.

Step 2: Solve for $t$ When $y = 0.25$

1. Substitute $y = 0.25$: $0.25 = \dfrac{40}{1 + 150e^{-0.4t}}$

2. Solve for $e^{-0.4t}$: $1 + 150e^{-0.4t} = \dfrac{40}{0.25} = 160$ $150e^{-0.4t} = 160 - 1 = 159$ $e^{-0.4t} = \dfrac{159}{150}$ $-0.4t = \ln\left( \dfrac{159}{150} \right)$ $t = -\dfrac{1}{0.4} \ln\left( \dfrac{159}{150} \right) \approx 12.263$

Step 3: Solve for $t$ When $y = 39.75$

1. Substitute $y = 39.75$: $39.75 = \dfrac{40}{1 + 150e^{-0.4t}}$

2. Solve for $e^{-0.4t}$: $1 + 150e^{-0.4t} = \dfrac{40}{39.75} \approx 1.00629$ $150e^{-0.4t} = 1.00629 - 1 = 0.00629$ $e^{-0.4t} = \dfrac{0.00629}{150}$ $-0.4t = \ln\left( \dfrac{0.00629}{150} \right)$ $t = -\dfrac{1}{0.4} \ln\left( \dfrac{0.00629}{150} \right) \approx 12.788$

Final Answer:

The news is spreading at a rate of 3.99 thousand people per day at approximately 12.263 days and 12.788 days.

(g) Finding the Fastest Rate at Which the News Spreads

Step 1: Identify the Maximum of $f'(t)$

• Objective: Find the maximum value of $f'(t)$.

Step 2: Recognize That Maximum Occurs at $y = \dfrac{M}{2}$

1. In logistic growth, the maximum rate occurs when $y = \dfrac{M}{2}$, where $M = 40$.

2. So, $y = 20$ thousand.

Step 3: Calculate the Maximum Rate

1. Use the differential equation: $f'(t) = 0.4y(40 - y)$

2. Substitute $y = 20$: $f'(t) = 0.4 \times 20 \times (40 - 20) = 0.4 \times 20 \times 20 = 160$

Final Answer:

The fastest rate at which the news spreads is 160 thousand people per day.

Understanding constraints

I’m analyzing the context of the problem, which involves optimizing $f(x, y) = 4x - 2y$ under the constraint $x^2 + 3y^2 = 156$.

Taking a closer look

I’m pulling together the problem by clearly defining what's given and what's needed. This involves understanding the concepts and equations to use and methodically solving each step to ensure accuracy throughout.

Outlining the approach

I’m working through the optimization problem by setting up equations with the gradients of the function and the constraint, using the method of Lagrange multipliers.

Pulling together gradients

First, I determined the gradients $\nabla f$ and $\nabla g$ to set up the Lagrange multiplier equations for optimization. This step involved solving straightforward algebraic expressions.

Linking equations

First, I’m connecting the derivatives, then substituting, and solving for $y$. This leads to:

Pinpointing solutions

OK, $x = -6y$ narrows down $y$ and gives us two points. Plugging these into the function, I find the maximum value is 52.

Understanding values

I’m summarizing the maximum and minimum values of a function, noting that the maximum is 52 and the minimum is -52.

Subject: Calculus - Optimization Using Lagrange Multipliers

Step 1: Given Information and Objective

• Given Function to Optimize:

  $$f(x, y) = 4x - 2y$$

• Constraint Equation:

  $$g(x, y) = x^2 + 3y^2 = 156$$

• Objective: Use Lagrange multipliers to find the maximum and minimum values of $f(x, y)$ subject to the given constraint.

Explanation:

We will use the method of Lagrange multipliers, which involves setting the gradient of the function equal to a scalar multiple of the gradient of the constraint:

where $\lambda$ is the Lagrange multiplier.

Step 2: Setting Up the Lagrange Multiplier Equations

Calculate the gradients:

• Gradient of $f$:

  $$\nabla f = \left( \dfrac{\partial f}{\partial x}, \dfrac{\partial f}{\partial y} \right) = (4, -2)$$

• Gradient of $g$:

  $$\nabla g = \left( \dfrac{\partial g}{\partial x}, \dfrac{\partial g}{\partial y} \right) = (2x, 6y)$$

Set up the equations:

1. $f_x = \lambda g_x$:

   $$4 = \lambda \cdot 2x$$

2. $f_y = \lambda g_y$:

   $$-2 = \lambda \cdot 6y$$

3. Constraint equation:

   $$x^2 + 3y^2 = 156$$

Step 3: Solving for $\lambda$ and Establishing a Relationship Between $x$ and $y$

From equation (1):

From equation (2):

Set Equations A and B equal to each other:

Cross-multiply:

Therefore:

Step 4: Substitute $x = -6y$ into the Constraint Equation

Substitute into $x^2 + 3y^2 = 156$:

Simplify:

Solve for $y^2$:

Step 5: Find Corresponding $x$ Values

When $y = 2$:

When $y = -2$:

Step 6: Calculate $f(x, y)$ at Each Point

First Point $(x, y) = (-12, 2)$:

Second Point $(x, y) = (12, -2)$:

Step 7: Determine the Maximum and Minimum Values

• Maximum Value:

  $$\text{Maximum} = 52 \quad \text{at} \quad (x, y) = (12, -2)$$

• Minimum Value:

  $$\text{Minimum} = -52 \quad \text{at} \quad (x, y) = (-12, 2)$$

Final Answer:

• The maximum value of $f(x, y)$ is 52.
• The minimum value of $f(x, y)$ is -52.