At the beginning of January 2010, Mr Wong’s fish farm has $40000$ fish breeding in a large po

Question

At the beginning of January 2010, Mr Wong’s fish farm has $40000$ fish breeding in a large pond. In the period between January and December each year, the number of fish increases by $9\%$. At the end of December each year, $5400$ fish are harvested and sold at the fish markets for $\$6$ per fish. No fish died or are being poached while he is in business. Find the number of fish just after the $n$ th harvest, giving your answer in the form $A-B{{\left( 1.09 \right)}^{n}}$, where $A$ and $B$ are constants to be determined.[3] Find the earliest month and year in which there will be no more fish in the pond.[2] Mr Wong has been thinking of retiring since January 2010. Every December just after the harvest, a nearby fish farm offers to buy over Mr Wong’s business by paying $\$8$ per fish for his remaining stock. If he wishes to retire by selling off his fish farm business after the $n$th harvest, find the value of $n$ where the total gains is the maximum.[3] White-tailed deer can be found in forests of Southern Canada. A wildlife biologist is investigating the change of a population of white-tailed deer of size $x$ (thousands) at time $t$ (years). He observed that on average, their birth rate is $4000$ per year and their death rate is proportional to the square of the population present. When the population was $5000$, the rate of increase of the population at that instant was $1750$ per year. Assuming that $x$ and $t$ are continuous variables, form a differential equation relating $x$ and $t$.[2] There were $2000$ deer in the forest initially. Solve this differential equation to obtain $x$ as an exact function of $t$.[4]

Accepted Answer

(a)(i) $A=60000$, $B=20000$ (a)(ii) December 2022 (a)(iii) $n=10$ (b)(i) $\dfrac{\mathrm{d}x}{\mathrm{d}t}=4-0.09x^2$ (b)(ii) $x=\dfrac{260e^{6t/5}-140}{21+39e^{6t/5}}$

2020 ASRJC P1 Q11

2020 ASRJC P1 Q11

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