Although you derivation takes on account both the Surface area and the Volume, I'd like to put this solution in prospective of the second term.
Since I do not know the value of the sides $\ell_{\mathrm{box}}$ an the volume $\ell_{\mathrm{box}}^3$, but I know his weight, then the only factor that correlates them is the Density, which I'll call $\varrho_x$, expressed by:
$$
\varrho_x=\frac{m_{\mathrm{box}}}{\ell_{\mathrm{box}}^3}
$$
Now, if the metal is the same, and so $\varrho_x$ doesn't change, then I could write a simple equation:
$$
\varrho_x=\frac{m_{\mathrm{box}(i)}}{\ell_{\mathrm{box}(i)}^3}=\frac{m_{\mathrm{box}(f)}}{(2\ell_{\mathrm{box}(i)})^3}
$$
The only unknown is $m_{\mathrm{box}(f)}$, so:
$$
m_{\mathrm{box}(f)}=\frac{m_{\mathrm{box}(i)}\cdot 8\ell_{\mathrm{box}}^3}{\ell_{\mathrm{box}(i)}^3}=m_{\mathrm{box}(i)}\cdot 8
$$
As you can see, the difference between the surface area is related to the difference of the volume, so you have to multiply by 8, as this simple $2^3$ times the side of the original lenght.
Let's take a generalized example.
If you imagine one simple cube, with lenght $a$, then the volume and the surface area are:
$$
S=6\cdot a^2,\quad V=a^3
$$
It's true that they have different factors, but they do not mean the same thing: if you imagine the increasing of weight, it's only given by the infinite variation of the Volume, as a simple integral:
$$
V_f=V_i+\left(\int_{a}^{2a}\mathrm{d}a\right)^3
$$
Above it can be observed that one single infintesimal change into the third dimension it's different from a single change of the surface; because the expansion is also inside the cube, which is not counted in the "$4$" factor, as two dimensional.
As you measure only the external surface, the volume takes the "internal expansion" into account, with either the relation of the density and the factor of $2$ "cubed" into the third dimension.
- There are two mixtures of wine and water, one of which contains twice as much water as wine, and the other three times as much wine as water. How much must there be taken from each to fill a pint cup, in which the water and wine shall be equally mixed?
Trying to solve this the same way as #26, we get the equations:
(27-1) $c + d = 1$
(27-2) $\frac{2c}{3} + \frac{d}{4} = \frac{1}{2}$
(27-3) $\frac{c}{3} + \frac{3d}{4} = \frac{1}{2}$
(27-4) $\frac{2c}{3} + \frac{d}{4} = \frac{c}{3} + \frac{3d}{4}$
Solving these equations gives an incorrect answer of $c = \frac{3}{5}$ and $d = \frac{2}{5}$.
To get the correct answer of $c = \frac{2}{3}$ and $d = \frac{1}{3}$ [...]
I got the numbers $c=\frac{2}{3}$ and $d=\frac{1}{3}$ by using the method mentioned at the end of the question's post. When I did this, I forgot to take into account that the first mixture has 3 parts (2 water, 1 wine) and the second has 4 parts (1 water, 3 wine). So, $c=\frac{2}{3}$, $d=\frac{1}{3}$ is actually wrong; $c=\frac{3}{5}$ and $d=\frac{2}{5}$ is right.
The issue wasn't that #26 and #27 couldn't be solved the same way, the issue was a careless mistake on my part.
Best Answer
HINT: