Compare the ratios under two sets of assumptions — the case when the side of the cube equals the radius of the cylinder, and the case when the side of the cube equals the diameter of the cylinder.
We will need to compute the ratios for cylindrical-shaped boxes and cube-shaped boxes using the appropriate formulas for surface area and volume. Let's start with a cylinder.
Ratio for Cylindrical-shaped Box
Let's look at the formulas for the volume and the surface area of a cylinder.
\begin{gathered}
\textcolor{deepskyblue}{V_\text{cylinder}}= \textcolor{darkviolet}{B} \cdot {\color{#FF0000}{h}} = \textcolor{darkviolet}{\pi r^2} \cdot {\color{#FF0000}{h}} = \textcolor{deepskyblue}{\pi r^2 h} \\
\quad \\
\textcolor{darkorange}{SA_\text{cylinder}}= \textcolor{darkorange}{2\pi r h+2\pi r^2}
\end{gathered}
We will substitute these values to find the ratio of the volume to surface area.
Let's now consider the formulas for the volume and the surface area of a cube.
\begin{gathered}
{\color{#0000FF}{V_\text{cube}}}={\color{#0000FF}{s^3}} \\
\quad \\
{\color{#009600}{SA_\text{cube}}}={\color{#009600}{6s^2}}
\end{gathered}
We will substitute these values to find the ratio of the volume to surface area.
To compare the efficiencies of the two boxes, we need to consider the ratios under the two cases as follows.
Radius of cylinder=Height of cylinder=Side of cube
Diameter of cylinder=Height of cylinder=Side of cube
Let's compare the ratios under the first case, where r=h=s.
Box Type
Ratio
Substitute r=h=s
Simplify
Value when r=1
Cylindrical
rh/2(r+h)
r( r)/2(r+ r)
r/4
0.25
Cube
s/6
r/6
r/6
0.17
By substituting the value of r=1 into our two ratios, we can see under the first set of assumptions that the cylindrical-shaped box has a larger ratio of surface area to volume. As a result, the cylindrical-shaped box will be more efficient. Let's repeat this process under the second case, 2r=h=s.
Box Type
Ratio
Substitute 2r=h=s
Simplify
Cylindrical
rh/2(r+h)
r( 2r)/2(r+ 2r)
r/3
Cube
s/6
2r/6
r/3
If 2r=h=s, the ratios of surface area to volume for both boxes are equal. As a result, the boxes would be equally efficient.
