A declining income against rising costs · source reference

Printed question

A fisherman purchases a boat for 1000 dollars at the beginning of 2011. Annual operating cost is 120 dollars in 2011 and increases by 40 dollars every year.

(i) Show total cost including purchase and operations through year $n$ is $20n^2+100n+1000$. [1]

Annual fishing income is 500 dollars in 2011 and decreases by 10% yearly due to overfishing.

(ii) Show cumulative fishing income never exceeds 5000 dollars, however many years pass. [2]

The boat can be sold for 600 dollars in 2011; sale price drops by 10% yearly. It is sold at the end of year $n$, after fishing income is collected.

(iii) Find total income from fishing and boat sale in terms of $n$. [3]

The fisherman sells in the final year when this total income exceeds total cost.

(iv) Find that year. [2]

Printed working

(i) Total cost up to year $n$:

$$1000+\frac n2[2(120)+(n-1)40]=1000+n(100+20n)=20n^2+100n+1000.$$

(ii) Fishing income up to year $n$:

$$\frac{500(1-0.9^n)}{1-0.9}<\frac{500(1-0)}{0.1}=5000\quad(n\in\mathbb Z^+).$$

It never exceeds 5000.

(iii) Income including sale at end of year $n$:

$$\frac{500(1-0.9^n)}{1-0.9}+600(0.9)^{n-1}=5000(1-0.9^n)+600(0.9)^{n-1}.$$

(iv) Consider

$$5000(1-0.9^n)+600(0.9)^{n-1}>20n^2+100n+1000.$$

GC table below shows the final profitable year is 2018.

Source tables and calculator displays

Term	Year	Operational cost (dollars)
1st	2011	120
2nd	2012	$120+40$
3rd	2013	$120+40(2)$
…	…	…
$n$th		$120+40(n-1)$

Term	Year	Fishing income (dollars)
1st	2011	500
2nd	2012	$500(0.9)$
3rd	2013	$500(0.9)^2$
…	…	…
$n$th		$500(0.9)^{n-1}$

Term	Year	Sale price (dollars)
1st	2011	600
2nd	2012	$600(0.9)$
3rd	2013	$600(0.9)^2$
…	…	…
$n$th		$600(0.9)^{n-1}$

Year	$n$	Total income minus total cost
2017	7	247.38 >0
2018	8	54.642 >0
2019	9	−198.8 <0

Printed answer key

(iii) $5000(1-0.9^n)+600(0.9)^{n-1}$; (iv) 2018

Mathematical review and qualifications

(i) $C_n=1000+\frac n2[240+40(n-1)]=20n^2+100n+1000$.

(ii) $I_n=500(1-0.9^n)/(1-0.9)=5000(1-0.9^n)<5000$.

(iii) Including sale gives $J_n=5000(1-0.9^n)+600(0.9)^{n-1}$.

(iv) $J_8-C_8\approx54.642>0$, while $J_9-C_9\approx-198.822<0$. Profit changes by $J_{n+1}-J_n-(C_{n+1}-C_n)=390(0.9)^{n-1}-(40n+120)$, which is strictly decreasing in n and already negative at n=8. Thus no later return to profit occurs. Last sale year is $2010+8=2018$.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: ARITHMETIC & GEOMETRIC PROGRESSIONS. [2014/CJC/JC1/mid-year/7] Question PDF pages 6; solution PDF pages 28, 29. The original files remain private QA records outside this native content.