Question

The fish population in a certain lake rises and falls according to the formula F = 3000(23 + 11t − t2). Here F is the number of fish at time t, where t is measured in years since January 1, 2006, 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, 2006? (b) By what date will all the fish in the lake have died? (Select the earliest date in which this occurs.)

Answer 1

Answer:

a) On January 1, 2017 the fish population will be the same as the initial population.

b) On September 18th, 2018 the fish population will be zero.

Step-by-step explanation:

Hi there!

a) First, let´s write the function:

F(t) = 3000(23 + 11t − t²)

The population on January 1, 2006 is the population at t = 0. Then:

F(0) = 3000(23 + 11· 0 - 0²)

F(0) = 3000 · 23 = 69000

This will be the population every time at which t² - 11t = 0. Then let´s find the other value of t (besides t = 0) that makes that expression to be zero:

t² - 11t = 0

t(t - 11) = 0

t = 0

and

t - 11 = 0

t = 11

On January 1, 2017 (2006 + 11), the fish population will be the same as the initial population.

b) We have to obtain the value of t at which F(t) = 0

F(t) = 3000(23 + 11t − t²)

0 = 3000(23 + 11t − t²)

divide both sides of the equation by 3000

0 = 23 + 11t − t²

Let´s solve this quadratic equation using the quadratic formula:

a = -1

b = 11

c = 23

x = [-b ± √(b² - 4ac)] / 2a

x = 12.8  ( the other value of x is negative and therefore discarded).

After 12.8 years all the fish in the lake will have died.

If 1 year is 12 months, 0.8 years will be:

0.8 years · 12 months/year = 9.6 months

If 1 month is 30 days, 0.6 month will be:

0.6 month · 30 days / month = 18 days

All the fish will have died after 12 years, 9 months and 18 days from January 1, 2006. That is, on September 18th, 2018.