67. A fish farmer has 5000 catfish in his pond. The number of catfish increases by $8 \%$ per month and the farmer harvests 300 catfish per month. (a) Show that the catfish population $P_{n}$ after $n$ months is given recursively by $$P_{n}-1.08 P_{n-1}-300 \quad P_{0}-5000 $$ (b) How many catfish are in the pond after six months?

Problem 67E

67. A fish farmer has 5000 catfish in his pond. The number of catfish increases by $8 \%$ per month and the farmer harvests 300 catfish per month.

(a) Show that the catfish population $P_{n}$ after $n$ months is given recursively by

$$P_{n}-1.08 P_{n-1}-300 \quad P_{0}-5000 $$

(b) How many catfish are in the pond after six months?

Step-by-Step Solution

(a)

To determine

To show: The population of catfish after n months is Pn=1.08Pn−1−300 where P0=5000 recursively.

Explanation

Given:

Number of catfish in the pond = 5000

Increase in percent of catfish = 8%

Number of fish harvest per month = 300

Proof:

The catfish population after 1 month is as follows.

P1=P0+8%P0−300=(1+0.08)P0−300=1.08P0−300

The catfish population after 2 month is as follows.

P2=P1+8%P1−300=(1+0.08)P1−300=1.08P1−300

The catfish population after 3 month is as follows.

P3=P2+8%P2−300=(1+0.08)P2−300=1.08P2−300

Proceed in similar way, the catfish population after n month is as follows:

Pn=Pn−1+8%Pn−1−300=(1+0.08)Pn−1−300=1.08Pn−1−300

Where, Pn−1 is the population of catfish at the end of the last month.

Therefore, it can be concluded that the population of catfish after n months is Pn=1.08Pn−1−300 where P0=5000 .

Hence the required proof is obtained.

(b)

To determine

To find: The number of catfish in the pond after six months.

Answer

The number of catfish in the pond after six months is 5734.

Explanation

Given:

The population of catfish after n months is Pn=1.08Pn−1−300 . (1)

Initial population, P0=5000 .

Calculation:

Obtain the value of P1 .

Substitute 1 for n in equation (1),

P1=1.08P1−1−300

P1=1.08P0−300 (2)

Substitute 5000 for P0 in equation (2),

P1=1.08(5000)−300=5400−300=5100

Thus, the value of P1=5100 .

Obtain the value of P2 .

Substitute 2 for n in equation (1),

P2=1.08P2−1−300

P2=1.08P1−300 (3)

Substitute 5100 for P1 in equation (3),

P2=1.08(5100)−300=5508−300=5208

Thus, the value of P2=5208 .

Obtain the value of P3 .

Substitute 3 for n in equation (1),

P3=1.08P3−1−300

P3=1.08P2−300 (4)

Substitute 5208 for P2 in equation (4),

P3=1.08(5208)−300=5624.64−300=5324.64

Thus, the value of P3=5325 .

Obtain the value of P4 .

Substitute 4 for n in equation (1),

P4=1.08P4−1−300

P4=1.08P3−300 (5)

Substitute 5325 for P3 in equation (4),

P4=1.08(5325)−300=5751−300=5451

Thus, the value of P4=5451 .

Obtain the value of P5 .

Substitute 5 for n in equation (1),

P5=1.08P5−1−300

P5=1.08P4−300 (6)

Substitute 5451 for P4 in equation (6),

P5=1.08(5451)−300=5887.08−300=5587.08

Thus, the value of P5=5587 .

Obtain the value of P6 .

Substitute 6 for n in equation (1),

P6=1.08P6−1−300

P6=1.08P5−300 (7)

Substitute 5587 for P5 in equation (6),

P6=1.08(5587)−300=6033.96−300=5733.96

Thus, the value of P6=5734 .

Therefore, the number of catfish in the pond after six months is 5734.