Answer
(a)$F=1000\times30=30,000$, so $t=17$(in 2019)
On January 1, 2019, the fish population again be the same as it was on January 1, 2002.
(b) $F=0$, so $t=18.61$ = $18$ years and $7.32$($0.61\times12$) months. So, the fish in the lake have died on July 2020.
Work Step by Step
$F=1000\times(30+17t-t^{2})$
(a) on January 1, 2002: $t=0$, $F=1000\times30=30,000$
The fish population again be the same as it was on January 1, 2002: $30,000=1000\times(30+17t-t^{2})$
$30,000=30,000+17000t-1000t^{2}$
$17000t-1000t^{2}=0$
$t=0$(in 2002) or $t=17$(in 2019)
So on January 1, 2019, the fish population again be the same as it was on January 1, 2002.
(b)All the fish in the lake have died when F=0: $30,000+17000t-1000t^{2}=0$
$-1000t^{2}+17000t+30,000=0$
$t=\frac{-17000+\sqrt (17000^{2}-4\times(-1000)\times30000)}{2\times(-1000)}$=$-1.61$ (reject)
or $t=\frac{-17000-\sqrt (17000^{2}-4\times(-1000)\times30000)}{2\times(-1000)}$=$18.61$
With $t=18.61=18$ years and $7.32$($0.61\times12$) months. So, the fish in the lake have died on July 2020.
