10.2 LINEAR GROWTH MODEL WORKSHEET SOLUTIONS
For each of the following linear growth model,
· Find an explicit description and the 9th generation of the population, P9
1) P0 = 718 and d = 32PN = 718 + 32N
P9 = 718 + 32(9) = 1006
2) 2, 19, 36, 53, …
d = 17 = 19 – 2
PN = 2 + 17N
P9 = 2 + 17 (9) = 155
3) 1123, 1332, 1541, 1750, …
d = 209 = 1332 – 1123
PN = 1123 + 209N
P9 = 1123 + 209(9) = 3004
4) 9.6, 10.05, 10.5, 10.95, …
d = 0.45 = 10.05 – 9.6
PN = 9.6 + .45N
P9 = 9.6 + .45(9) = 13.65 / 5) P5 = 71 and P6 = 84
d = 84 – 71 = 13; P0 = 71 – 13(5) = 6
PN = 6+ 13N
P9 = 6 + 13(9) = 123
6) P13 = 96 and P14 = 105
d = 105 – 96 = 9; P0 = 96 – 9(13) = -21
PN = -21 + 9N
P9 = -21+ 9(9) = 60
7) P6 = 65 and P15 = 128
d = (128 – 65)/(15-6) = 7;
P0 = 65 – 7(6) = 23
PN = 23 + 7N
P9 = 23 + 7(9) = 86
8) P21 = 2170 and P90 =5896
d = (5896 – 2170)/(90-21) = 54;
P0 = 2170 – 54(21) = 1036
PN = 1036 + 54N
P9 = 1036 + 54(9) = 1522
9) Every 3 months, Marcus buys himself 4 new neckties. Let P0 represent the number of neckties he starts with and PN be the number of neckties in his collection at the end of Nth 3-month period. Assume that he started out with 3 neckties and he never throws away one.
a. What is the population in this problem? Neckties
b. What is the transition? 3 – months
c. What is the transition rule? PN = PN-1 + 4; Adding 4 neckties
10) A population of lab rats is going to be increased by 3 rats a month. If it costs $3.50 to care for each rat a month and there were 2 rats to begin with in the lab, then how much will the lab be paying for the rats after 10 months? Assume all rats survive lab work.
P10 = 2 + 3(10) = 32 rats; 32(3.50) = $112
For each of the following linear growth model, find the given sum
11) 5 + 8 + 11 + 14 + 17 + 20 + 23 + 26 + 29= (5 + 29)*9/2 = 153
12) P0 = 115 and P29 = 318;
P0 + P1 + P2 + … + P29
= (115 + 318)*30/2 = 6495
13) For 50 terms, 29 + 44 +59 +…
P49 = 29 + 15(49) = 764
= (29 + 764)*50/2 = 19,825
14) For 125 terms, 421 + 427 + 433 + …
P124 = 421 + 6(124) =1165
(421 + 1165)*125/2 = 99,125 / 15) P0 = 111 and d = 31;
P0 + P1 + … + P254
P254 = 111 + 31(254) =7985
(111 + 7985)*255/2 = 1,032,240
16) P0 = 12 and d = 4.5;
P50 + P51 + … + P119
P50 = 12 + 4.5(50) = 237
P119 = 12 + 4.5(119) = 547.5
(237 + 547.5)*70/2 = 27,457.5
17) 5 + 17 + 29 + 41 + … + 1157
1157 = 5 + 12N; N = 96
(5 + 1157)*97/2 = 56,357
18) 9012 + 8895 + 8778 + … + 7140
7140 = 9012 – 117N; N = 16
(9012 + 7140)*17/ 2 = 137,292
19) A small business sells $6,500 worth of products during its first year of business. The owner has set an annual goal for increased sales at $3500 for 15 years. Find the total sales during the first 8 years this business is in operation. N = 0 à First Year, N = 7 à 8th year
P7 = 6500 + 3500(7) = 31,000
(6500 + 31,00)*8/2 = $150,000
20) Cosmic Java opened in 2005 with 21 coffee shops and plans to open 13 new coffee shops each year after that. Suppose it costs the company $45,000 each year to maintain one coffee shop.
a. How many Cosmic Java coffee shops will there be in 2020?
N = 0 à 2005, 2020 à N = 15 21 + 13(15) = 216 SHOPS
b. In what year will Cosmic Java reach the milestone of having 300 coffee shops?
300 = 21 + 13N; N = 21.5 or between 2026 and 2027
c. In 2010, how much money will the company have used to maintain all its coffee shops since 2005?
P5 = 21 + 13(5) = 86; (21 + 86)*6/2 = 321 Shops;
321*45000 = $14,445,000