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.
