Suppose we have a population of fish, say $10000$, with an exponential growth each year of $30\%$.
If we want to collect as many fish as possible in, say 10 years, a natural question to ask is:
What is the optimal number of fish to catch each year?
Mathematically speaking, we have a sequence defined by:
$a_{n+1}=1.3 \cdot (1-k_n)a_n$
$a_0=10000$
where $k_n$ is the percentage of fish we want to collect in the $n$th year. We want to choose $k_1, k_2, k_3, \dots, k_n$ to maximize $a_{10}$.
Also, $k$ is capped at some number. Let it be $k \leq 0.05$.
I think this problem is quite economically relevant, but can't think of any optimal solution for it.
Instead of deciding how much to fish, let us decide how much not to fish. Let $x_k$ be the number of fish at year $k$ and let $u_k \in [0.95,1]$ be the percentage of fish not to fish at year $k$. Hence,
$$x_{k+1} = 1.3 \, x_k \, u_k $$
and
$$\begin{array}{rl} x_1 &= 1.3 \, x_0 \, u_0\\ x_2 &= 1.3 \, x_1 \, u_1 = 1.3^2 \, x_0 \, u_1 \, u_0\\ x_3 &= 1.3 \, x_2 \, u_2 = 1.3^3 \, x_0 \, u_2 \, u_1 \, u_0\\ &\vdots\\ x_N &= 1.3^N \, x_0 \displaystyle\prod_{k=0}^{N-1} u_k\end{array}$$
As $x_0$ is given, maximizing $x_N$ is equivalent to maximizing the product $\displaystyle\prod_{k=0}^{N-1} u_k$. Hence,
$$u_0 = u_1 = \cdots = u_{N-1} = 1$$
In other words, the optimal fishing strategy is not to fish at all! It would probably be wise to update the objective function.