I have been given the basic differential equation for population modelling, which is $$p'(t)=rp(t)\left(1-\frac{p(t)}k\right)$$. The coefficient $r$ is $10$ and constant $k$ is $500$, so the carrying capacity is $500$ fish. Each fish causes $5$ dollars of expenses per year, and the fish tank costs $1200$ dollars per year to operate. Each sold fish brings $15$ dollars of revenue.
I have been given the task of finding out the best way to harvest the fish without destroying the population and also with highest possible revenue. The assignment says that you can either harvest the fish certain fixed number evey year or harvest the fish based on the population with a certain multiplier.
Assuming that we use the constant amount $700$ and the multiplier $5$, we can get two different equations that model the fish population absed on two different strategies.
$$p'(t)= rp(t)\left(1-\frac{p(t)}k\right)-C$$ ($C$ being constant)
$$p'(t)=rp(t)\left(1-\frac{p(t)}k\right)-Bp(t)$$ ($B$ being a multiplier)
When we include the numerical values, we get:
$$p'(t)= 10p(t)\left(1-\frac{p(t)}{500}\right)-700\\ p'(t)=10p(t)\left(1-\frac{p(t)}{500}\right)-5p(t)$$
So far so good. But my problem is that how am I supposed to make a differential function that shows the rate of revenue? My attempt was the following:
$$v'(t)=15 \cdot 700-5p(t)-1200\\ v'(t)= 15 \cdot p(t)-5p(t)-1200$$
My other attempt:
$$v'(t)= 15 \cdot 700t-5p(t)-1200t\\ v'(t)= 15 \cdot 5p(t)-5p(t)-1200t$$
I am making this with matlab, but the basic calculations are the thing that bugs me. Have I even done the first parts correctly? How can I form the revenue functions so that they are consistent? Any help would be appreciated.
$v'(t) = \frac {dv}{dt}$ is how much $v(t)$ changes each year. I.e., it is how much profit you make in year $t$.
Consider the constant-sell strategy. In a given year, you sell $C$ fish. That brings in $15C$ in income. You spend $5p(t)$ on upkeep of the fish and $1200$ on upkeep of the tank. So your profit is income less outgo: $$v'(t) = (15C) - (5p(t) + 1200)$$
In your second attempt, you get the same result in year one, but in year $t = 2$, you are claiming to bring in $15C\cdot 2 = 30C$ in income, and $15C\cdot 3 = 45C$ in year $3$. Seems a pretty sweet deal, though the equally rising tank maintenance fees are taking a chunk out of it. At least the cost of feeding the fish isn't also rising each year.