ONE Maths Word problem

No. of fish by December 1998 = 100,000 (1 + 10%) = 110,000.
After selling 15,400, fish remaining for January 1999 = 110,000 - 15,400 (this is the quantity after the 1st harvest, n = 1)
(a) No. of fish by December 1999 = (110,000 - 15,400) (1 + 10%)
= 110,000 (1.1) - 15,400 (1.1) = 104,060.
(b) After selling 15,400, fish remaining for January 2000
= 110,000(1.1) - 15,400(1.1) - 15,400 (this is the quantity after the 2nd harvest, n = 2)
No. of fish by December 2000 = [ 110,000(1.1) - 15,400 (1.1) - 15,400] (1 + 10%)
= 110,000(1.1)^2 - 15,400(1.1)^2 - 15,400(1.1) = 97526
After selling 15,400, fish remaining for January 2001
= 110,000(1.1)^2 - 15,400(1.1)^2 - 15,400(1.1) - 15,400 (this is the quantity after the 3rd harvest, n = 3).
From here you can see that quantity after n harvest Fn
= 110,000(1.1)^(n - 1) - 15,400(1.1)^(n - 1) - 15,400(1.1)^(n - 2) - ..... - 15,400(1.1) - 15,400
= 110,000(1.1)^(n - 1) - 15,400[ 1 + 1.1 + 1.1^2 + 1.1^3 + .... + 1.1^(n - 1)]
= 110,000(1.1)^n/1.1 - 15,400 (1.1^n - 1)/(1.1 - 1)
= 100,000(1.1)^n - 154,000(1.1)^n + 154,000
= (100,000 - 154,000)(1.1)^n + 154,000
= 154,000 - 54,000(1.1)^n.
(c)
When all fish are sold Fn = 0
so 54,000(1.1)^n = 154,000
1.1^n = 154000/54000 = 77/27
n = log (77/27)/log 1.1 = 10.995
so all fish are sold after the 11 harvest = 1999 + 11 = January 2010.
For (d), see if I can find the easiest method and advise later.


2014-09-27 17:23:24 補充：
(d) Total income is given by : [154,000 - 54,000(1.1)^n + 15,400](15) + 15,400(10)(n - 1). This is an exponential function that is not easy to find the max value even by calculus. So try n = 4,5,6,.... is the only method. Max is when n = 7. Total income is 1,886,539. Farmer should sell in 2005.