Optimise the number of coins a boat can hold

Forces acting upon the boat

Setup. In order for the boat to remain in equilibrium and to maximise the number of coins, it must be (almost) fully submerged in water. Therefore, we can claim that $\vec{B}+\vec{W}=0, \text{ where }B\text { is buoyancy and } W=W_b+W_c\text{, the weight of the boat + the weight of the coins}$

Expressing each term individually, we get $\rho Vg=m_bg+Nmg\implies N=\left\lfloor\frac{\rho V-m_b}{m}\right\rfloor$

The mass of the boat is given by $m_b=\sigma L^2$ , because we are given the areal density of the paper, and we're left with $N=\left\lfloor\frac{\rho V-\sigma L^2}{m}\right\rfloor\tag{1}$

From this expression, we can assess that as $V$ grows, $N$ must grow too. So it's all about maximising the volume!

When is the volume maximal? First, let's express the volume as a function of two parameters, $L\text{ and }h$ . Looking at the scheme in the problem body, it is easy to establish that

$V(h, L)=(L-2h)^2h=4h^3-4Lh^2+L^2h$

Now, we know that a derivative expresses the rate of change of a function's output for certain values of its parameters. Keeping that in mind, we can deduce that when a function reaches its maximal value, its derivative must be $0$ , as it cannot increase any further. However , this also happens when the function reaches its minimal value, as it cannot decrease any further. That being said, we aim to compute the derivative of $V$ with respect to $h$ , and setting it to be $0$ , then being careful about which of the solutions of the obtained polynomial fulfils the necessary criteria for the volume to be maximal.

$\frac{\partial V}{\partial h}=\frac{\partial}{\partial h}\left(4h^3-4Lh^2+L^2h\right)$ $\frac{\partial V}{\partial h}=\frac{\partial}{\partial h}\left(4h^3\right)+\frac{\partial}{\partial h}\left(-4Lh^2\right)+\frac{\partial}{\partial h}\left(L^2h\right)$ $\frac{\partial V}{\partial h}=12h^2-8Lh+L^2$

Imposing the extremum condition, then solving the quadratic yields:

$\frac{\partial V}{\partial h}=0\implies 12h^2-8Lh+L^2=0\implies h\in \left\{\frac L 2, \frac L 6\right\}$

But $\frac L 2$ is definitely not the solution we were seeking, as the volume in that case would be $0$ , so we're left with

$h=\frac L 6\implies V=\frac{2L^3}{27}\tag{2}$

From relations $(1)\text{ and }(2)$ , we can infer that

$N=\left\lfloor\frac{\rho \frac{2L^3}{27}-\sigma L^2}{m}\right\rfloor$ $N=\left\lfloor \frac{L^2}{m}\left(\frac{2\rho L}{27}-\sigma\right) \right\rfloor$

Calculating with the given values, $N\approx \lfloor 1841.85\rfloor$ , and rounded to the nearest multiple of $10$ , we get our answer, $\boxed{1840}$ .