Question 25
A lake in Northern Ontario was overfished, reducing its fish population. It was restocked with 800 fish three years ago; the current population is estimated at 6000 and the carrying capacity is 8000. From the restocking time, model population by $P(t)=c/(1+a b^{-t})$, with real parameters $a,b,c$ and $t$ measured in years, as in the graph. Let $g(t)$ be its growth rate. The supplied ALGORITHM 1 initializes $a=0$, $b=10$, $\Delta=0.05$, $\mathrm{max\_t}=a$, $\mathrm{max\_g}=g(a)$, $t=a$; while $t\le b$, compute $\mathrm{value}=g(t)$, update the stored maximum and its time if value is larger, and increase $t$ by $\Delta$. Output the stored maximum and time. (The algorithm's endpoint variable names reuse the model's letters.) Using this algorithm, the value $g(t)$ at the 41st step is $A$. The actual maximum growth rate found mathematically is $B$. Find $B-A$, rounding all results to hundredths.
The following source mathematical images preserve the printed option/formula/figure variants; text and choices above or below provide clean typesetting.
The following source mathematical images preserve the printed option/formula/figure variants; text and choices above or below provide clean typesetting.