Chapter 1.5, Problem 115E

Fish Population The fish population in a certain lake rises and falls according to the formula

$F=1000(30+17t-t^2)$

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.

1. (a) On what date will the fish population again be the same as it was on January 1, 2002?
2. (b) By what date will all the fish in the lake have died?

To determine

The date on which the fish population will be the same as it was on January 1, 2002.

Answer to Problem 115E

The fish population will be the same as it was on January 1, 2002 after $\underset{\_}{17\text{years}\text{on}\text{Jan}\text{1,}\text{2019}}$.

Explanation of Solution

Given:

The fish population in a certain lake rises and falls according to the formula $F=1000\left(30+17t-t^2\right)$, where $F$ is the number of fish at time $t$ and $t$ is measured in years since January 1, 2020.

Calculation:

Substitute $0$ for $t$ in the equation $F=1000\left(30+17t-t^2\right)$, to obtain the date on which the fish population will be the same as it was on January 1, 2002.

$\begin{array}{c}F=1000\left(30+17\left(0\right)-{\left(0\right)}^2\right)\\ F=1000\times 30\\ F=30000\end{array}$

Therefore, the initial fish population is $30000$.

Substitute $30000$ for $F$ in the equation $F=1000\left(30+17t-t^2\right)$.

$\begin{array}{l}30000=1000\left(30+17t-t^2\right)\\ 30+17t-t^2=\frac{30000}{1000}\\ 17t-t^2=30-30\end{array}$

Simplify the above equation as follows.

$\begin{array}{l}17t-t^2=0\\ t\left(17-t\right)=0\\ t=0\text{or}17-t=0\\ t=0\text{or}t=17\end{array}$

So, the population of fish will be same again as it was on $\text{January}\text{1,}\text{2002}$ after $\text{17}\text{years}\text{on}\text{Jan}\text{1,}\text{2019}$.

Thus, the fish population will be the same as it was on January 1, 2002 after $\underset{\_}{17\text{years}\text{on}\text{Jan}\text{1,}\text{2019}}$.

To determine

The date by which all the fish in the lake will be dead.

Answer to Problem 115E

All the fish in the lake will be dead after $\underset{\_}{18.612\text{years}\text{on}\text{Aug}\text{12,}\text{2020}}$.

Explanation of Solution

Formula used:

Quadratic formula:

The solution of a quadratic equation of the form $a{x}^2+bx+c=0,a\ne 0$ can be obtained by using the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.

Calculation:

The number of fish will be zero when all the fish in the lake are dead.

Therefore, substitute $0$ for $F$ in the equation $F=1000\left(30+17t-t^2\right)$.

$\begin{array}{l}0=1000\left(30+17t-t^2\right)\\ t^2-17t-30=0\end{array}$

Use Quadratic formula to find the value of t.

Substitute $1$ for $a$, $-17$ for $b$ and $-30$ for $c$ in the quadratic formula.

$\begin{array}{c}t=\frac{-\left(-17\right)\pm \sqrt{{\left(-17\right)}^2-4\times 1\times \left(-30\right)}}{2\times 1}\\ =\frac{17\pm \sqrt{289+120}}{2}\\ =\frac{17\pm \sqrt{409}}{2}\end{array}$

Simplify the above equation as follows.

$\begin{array}{c}t=\frac{17\pm 20.22}{2}\\ t=\frac{17+20.22}{2}\text{or }t=\frac{17-20.22}{2}\\ t=\frac{37.22}{2}\text{or }-\frac{3.22}{2}\\ t=18.612\text{or }-1.61\end{array}$

The number of years must be positive.

Therefore, all the fish in lake will be dead after $18.612\text{years}$.

Assuming that there will be 365 days in a year, convert 0.612 years to days.

$0.612\text{years}\times 366=224\text{days}$

That is, all the fish in lake will be dead after $18\text{years}\text{and 224 days}$.

By using a calendar, it is obtained that all the fish will be dead on $\text{Aug}\text{12,}\text{2020}$.

Thus, all the fish in the lake will be dead after $\underset{\_}{18.612\text{years}\text{on}\text{Aug}\text{12,}\text{2020}}$.