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

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

Here $F$ is the number of fish at the time $t$, where $t$ is measured in years since January 1, 2002, when the fish population was first estimated.

On what date will the fish population again be the same as it was on January 1, 2002?

By what date will all the fish in the lake have died?

  • I don't know exactly how to go about solving this problem. I suspect that it requires for me to solve for one variable in terms of another. I think that it wants me to solve for $t$ in terms of $F$

So this is what I got:

$$F=30000+17000t-1000t^2$$ $$F-30000=17000t-1000t^2$$

Am I going in the right direction?

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    $\begingroup$ Measure population in thousands. Then we can forget about the $1000$ in front. The population at time $0$ is $30$ (thousand). When is $30+17t-t^2=30$? $\endgroup$
    – André Nicolas
    Commented Aug 23, 2014 at 16:09
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    $\begingroup$ As to whether you are going in the right direction, the manipulations you did are correct. However, it is not clear what the intended next step is. $\endgroup$
    – André Nicolas
    Commented Aug 23, 2014 at 16:18
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    $\begingroup$ For the first question, you are solving $30=30+17t=t^2$, which simplifies to $17t-t^2=0$. We need nothing fancy for this, for we are looking at $t(17-t)$, which is $0$ at the beginning and also when $t=17$. For the second question, you will need to solve $30+17t-t^2=0$. To solve this you will need the Quadratic Formula. $\endgroup$
    – André Nicolas
    Commented Aug 23, 2014 at 16:35
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    $\begingroup$ At $t=0$ the population is $30$ (thousand). The question asks at what time $t$ the population will be the same as at the beginning, i.e. $30$. To answer this, we set $30+17t-t^2=30$ and solve for $t$. $\endgroup$
    – André Nicolas
    Commented Aug 23, 2014 at 16:52
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    $\begingroup$ It says that the population is given by a certain formula, and that time is measured from Jan 1, 2002. So at the beginning, we have $t=0$. Plug in $0$ for $t$ in the given formula. $\endgroup$
    – André Nicolas
    Commented Aug 23, 2014 at 17:06

3 Answers 3

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First, use the equation you were given to determine what the population $F$ was in $2002$, when $t=0$. Then the first question asks you to determine when the population will again be that much. Suppose the answer you get for the 2002 population is $F_0$. Then you want to solve $$F_0 = 1000(30+17t-t^2)$$ or $$1000t^2 - 17000t + (F_0-30000) = 0.$$ This is a quadratic equation; I suspect you know how to solve it.

For the second question, you want to solve $0 = 1000(30+17t-t^2)$ to find when the population is zero. Divide both sides by $1000$ to simplify, again giving a quadratic that you can solve for $t$.

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  • $\begingroup$ @VarunIyer I know that. I was trying to lead the OP through the solution rather than telling him/her the answer. $\endgroup$
    – rogerl
    Commented Aug 23, 2014 at 16:12
  • $\begingroup$ @VarunIyer there's nothing wrong with his method. $\endgroup$
    – candido
    Commented Aug 23, 2014 at 16:13
  • $\begingroup$ The OP asked two questions: the first was when the population is again the same as in 2002; that is what I answered in my first paragraph. His second question was when the population will have died out. This is what I answered in my second paragraph. $\endgroup$
    – rogerl
    Commented Aug 23, 2014 at 16:14
  • $\begingroup$ Were my manipulations correct or unnecessary? If they were correct, I have no idea what the next step should be in the process. $\endgroup$
    – Cherry_Developer
    Commented Aug 23, 2014 at 16:33
  • $\begingroup$ The first, writing $F = 30000 + 17000t - 1000t^2$, is just fine. The second doesn't really accomplish much. If you try to focus on what data you are given ($F$ in 2002, or $F=0$), and what you have to find ($t$), you see that you can really think of this as an equation in one unknown, $t$, and rewrite it appropriately. From your first equation, I would proceed to say $1000t^2 - 17000t + (F-30000) = 0$ and then ask: for the first question, I want to solve this quadratic for $F=30000$; for the second, at $F=0$. Hope that helps. $\endgroup$
    – rogerl
    Commented Aug 23, 2014 at 18:37
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Yes you are.

So,

notice that initially:

$$F(0) = 30000$$

So, we must find a time where the population equals $30,000$

So,

$$F=30000+17000t-1000t^2 = 30000$$

Simple algebra from here:

$$17000t - 1000t^2 = 0$$ $$1000t(17 - t) = 0$$

So $t = 17$

Comment if you have any questions.

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  • $\begingroup$ I am really just wondering how this all has to be manipulated to arrive at the answer. $\endgroup$
    – Cherry_Developer
    Commented Aug 23, 2014 at 16:41
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First you should find out how many fish there were in the lake on January 1, 2002 (i.e., t=0). This will give you a value for F, say F=a. Then the question becomes, for what other values of t does F=a?

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