A deeper beam bends less

A wooden beam $3.0\ \text{m}$ long and $8.0\ \text{cm}$ wide rests on two supports and carries $4.0\ \text{kN}$ at its middle. How deep must it be? You will make the program find the sag for every depth from $5\ \text{cm}$ to $30\ \text{cm}$, the smallest depth that keeps the sag under $5\ \text{mm}$, and how much less a beam twice as deep sags.

The idea. A beam loaded at its middle sags $\delta = \dfrac{Wl^3}{4bd^3Y}$. For wood, $Y$ is about $1.0 \times 10^{10}\ \text{Pa}$.

Your tasks.

1. Work out the sag at every depth in d.
2. Find the smallest depth whose sag is at most $5\ \text{mm}$.
3. Find how many times the sag at $10\ \text{cm}$ is the sag at $20\ \text{cm}$.