Chapter 8.1, Problem 43E

a.

To determine

To Show : That the catfish population P_nafter n months is given by, $P_n=1.08P_{n-1}-300$ .

Answer to Problem 43E

The catfish’s population P_nafter n months is given by, $P_n=1.08P_{n-1}-300,P_0=5000$ .

Explanation of Solution

Given Information :

The fish farmer has 5000 number of catfish in the pond. That is, P₀= 5000 . The number of catfish increases at a rate of 8% per month and the farmer harvests 300 catfish each month.

P₀=5000, that is initially the farmer has 5000 catfish. $p_1=(5000-300)+8\%\text{increase}\text{in}\text{the}\text{number}\text{of}\text{catfish}$ .

Thus,

The number of balance catfish in the n^thmonth is, $P_n=(1+.08)P_{n-1}-300$

That is,

  $P_n=1.08P_{n-1}-300$ .

b.

To determine

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

Answer to Problem 43E

The number of catfish in the pond after six months is, $P_6=5730$ .

Explanation of Solution

Calculation:

As shown in part a,

The number of catfish present in the pond after n months is, $P_n=1.08P_{n-1}-300$ .

Therefore we get,

  $P_1=(1.08)P_0-300$

  $=(1.08\times 5000)-300=5100$

  $P_2=(1.08)P_1-300=(1.08\times 5100)-300=5208$

  $P_3=(1.08)P_2-300=(1.08\times 5208)-300=5324$

  $P_4=(1.08)P_3-300=(1.08\times 5324)-300=5449$

  $P_5=(1.08)P_4-300=(1.08\times 5449)-300=5584$

  $P_6=(1.08)P_5-300=(1.08\times 5584)-300=5730$

Conclusion:

Thus, the number of catfish present in the pond after six months is 5730.