# Solving for $\frac{dP}{dt}=aP\left(1-\frac{P}{b}\right)-\left(\frac{c}{100}\right)P=0$

A fish farm grows and harvests barramundi in a large dam. The population of fish after $t$ years is given by the function $P(t)$. The rate of change in the population $\frac{dP}{dt}$ is modeled by $\frac{dP}{dt}=aP\left(1-\frac{P}{b}\right)-\left(\frac{c}{100}\right)P$ where $a$, $b$ and $c$ are known constants. $a$ is the birth rate of the barramundi, $b$ is the maximum carrying capacity of the dam and $c$ is the percentage that is harvested each year.

If the birth rate is 6%, the maximum carrying capacity is 24 000 , and 5% is harvested each year, find the stable population.

So basically what the question is asking for is when is $\frac{dP}{dt}=0$. However, when I plug in the constants and solve for $P$, I get the following: $$0.06P\left(1-\frac{P}{24000}\right)-\left(\frac{0.05}{100}\right)P=0$$ $$0.06P-\frac{0.06P^2}{24000}=\frac{0.05P}{100}$$ $$6P-\frac{6P^2}{24000}=0.05P$$ $$\frac{6P^2}{24000}-5.95P=0$$ $$P\left(\frac{6P}{24000}-5.95\right)=0$$ $$P=0 \ \text{or} \ P=23800$$ While the book gets $P=4000$. Could someone put me on the right track to the correct answer? Maybe I misinterpreted the question. Thanks in advance!

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Minor error. The harvest rate is $5$ percent, so you should have had $\dfrac{5}{100}$, not $\dfrac{0.05}{100}$.
Remark: In some situations, a case could be made for your interpretation, since technically $5$ percent means $\frac{5}{100}$, and the equation seems to ask for a further division by $100$. However, the context shows that this is not what is intended. Harvesting $5$% of the current population $P$ means decreasing the population by $\frac{5}{100}P$, or equivalently by $(0.05)P$.