Question

The fish population in a certain lake rises and falls according to the formula F = 4000(26 + 12t − 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.

(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?

Answer 1

The fish population will be the same as it was on January 1, 2002, on January 1, 2007. The fish population will be extinct by approximately 6 + sqrt(62) years after January 1, 2002.

Step-by-step explanation:

To find the date when the fish population will be the same as it was on January 1, 2002, we need to set F equal to the initial population, which is 4000. So, we have:

4000 = 4000(26 + 12t - t^2)

Simplifying:

1 = 26 + 12t - t^2

Rearranging the equation:

t^2 - 12t + 25 = 0

Factoring:

(t-5)(t-5) = 0

t = 5

Therefore, the fish population will be the same as it was on January 1, 2002, on the date 5 years later, which is January 1, 2007.

To find the date when all the fish in the lake have died, we need to set F equal to 0. So, we have:

0 = 4000(26 + 12t - t^2)

Solving for t:

t^2 - 12t - 26 = 0

Using the quadratic formula:

t = (12 ± sqrt(12^2 - 4(1)(-26))) / 2(1)

Simplifying:

t = (12 ± sqrt(144 + 104)) / 2

t = (12 ± sqrt(248)) / 2

t = (12 ± 2sqrt(62)) / 2

t = 6 ± sqrt(62)

Therefore, the fish population will be extinct by the date approximately 6 + sqrt(62) years after January 1, 2002.