Problem 99 - Algebra and Trigonometry

Question

Fish Population The fish population in a certain lake rises and falls according to the formula $$ F=1000\left(30+17 t-t^{2}\right) $$ Here \(F\) is the number of fish at time \(t\) , where \(t\) is measured in years since January \(1,2002\) , when the fish population was first estimated. (a) On what date will the fish population again be the same as it was on January \(1,2002 ?\) (b) By what date will all the fish in the lake have died?

Step-by-Step Solution

Answer

(a) January 1, 2019; (b) early July 2020.

Step 1: Determine Initial Fish Population

First, calculate the initial fish population when \( t = 0 \), which corresponds to January 1, 2002. Substitute \( t = 0 \) into the formula: \[ F = 1000(30 + 17 \, \cdot \, 0 - 0^2) = 1000 \, \cdot \, 30 = 30000. \] So, the initial population is 30,000 fish.

Step 2: Equate Formula to Initial Population

To find when the fish population returns to 30,000, set \( F = 30000 \) and solve for \( t \). The equation becomes: \[ 1000(30 + 17t - t^2) = 30000. \] Simplify to: \[ 30 + 17t - t^2 = 30. \]

Step 3: Solve for Time When Population Returns to Initial Level

The equation from Step 2 simplifies to: \[ 17t - t^2 = 0. \] Factor the quadratic equation: \[ t(17 - t) = 0. \] This gives solutions \( t = 0 \) and \( t = 17 \). Since \( t = 0 \) is the initial time, the fish population will be the same again at \( t = 17 \).

Step 4: Calculate Time Until Fish Die Out

For all fish to die, set \( F = 0 \) and solve \( 1000(30 + 17t - t^2) = 0 \). Simplify to: \[ 30 + 17t - t^2 = 0. \] Rearrange into standard form: \( t^2 - 17t - 30 = 0 \).

Step 5: Solve Quadratic for Extinction Time

Use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to solve \( t^2 - 17t - 30 = 0 \) where \( a = 1 \), \( b = -17 \), \( c = -30 \). Calculate the discriminant: \[ b^2 - 4ac = 289 + 120 = 409. \] Thus, \[ t = \frac{17 \pm \sqrt{409}}{2}. \] Calculate: \( t \approx \frac{17 \pm 20.19}{2} \). The solutions for \( t \) are: \( t \approx 18.595 \) and \( t = -1.595 \). Only positive \( t \approx 18.595 \) is valid.

Step 6: Determine the Calendar Dates

The fish population will return to the initial level on January 1, 2002 plus 17 years, which is January 1, 2019. The fish population will die out approximately 18.595 years from January 1, 2002, which calculates to early July 2020.

Key Concepts

Quadratic Equation

Understanding the fish population model involves working with a type of mathematical expression known as a quadratic equation. Quadratic equations are polynomial equations of the second degree, typically written in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. In the given exercise, the fish population model is expressed as \( F = 1000(30 + 17t - t^2) \). This equation models how the population changes over time, based on the values of time \( t \).
Quadratic equations can have two solutions, which means the population model could show two critical points. Finding these points is essential for understanding the behavior of the fish population over time.

Initial Population Calculation

To determine the initial fish population on January 1, 2002, we evaluate the given quadratic model at \( t = 0 \). By substituting \( t = 0 \) into the equation \( F = 1000(30 + 17t - t^2) \), we simplify to \( F = 1000(30) \), which equates to a population of 30,000 fish.
This value represents the starting point for our study of the fish population over time and provides a benchmark from which we can measure changes. Knowing the initial population is vital for answering later parts of the exercise, such as when the population returns to its original size or drops to zero.

Extinction Time

Finding the time it takes for the fish population to go extinct requires solving the equation for when \( F = 0 \). This extinction is calculated by setting the population equation \( 1000(30 + 17t - t^2) = 0 \), leading to the quadratic \( 30 + 17t - t^2 = 0 \). By rearranging, we form the standard quadratic \( t^2 - 17t - 30 = 0 \).
Solving this equation gives us potential times \( t \) where the population could become zero. Only the positive solution is physically meaningful, since time cannot be negative, providing an estimate for when the lake's fish population will disappear. In this case, it occurs approximately 18.595 years after the initial assessment.

Quadratic Formula

The quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) is a reliable tool for solving any quadratic equation. For the exercise at hand, the quadratic \( t^2 - 17t - 30 = 0 \) is solved using this method, where \( a = 1 \), \( b = -17 \), and \( c = -30 \).
First, the discriminant \( b^2 - 4ac \) is calculated, which helps to determine the nature of the roots. With a positive discriminant of 409, there are two distinct real roots. Plugging these values into the formula provides solutions for \( t \). For practical purposes in this problem, we only consider the positive solution, which helps us determine the fish population's time to extinction, approximately 18.595 years from the start date.