Chapter 8.4, Problem 31PPSE

Solution

To identify: The comparison of ratios of the volume to the surface area of the containers, how the cereal’s box efficiency is measured, and the formula used to solve the problem if a cereal company is considering a cylindrical shaped box and cube shaped box for packaging.

The cereal’s box efficiency can be measured by comparing their ratio of volume to surface area and assuming different scenarios of radius, height, and side of cube.

Given information:

A cereal company is considering a cylindrical shaped box and cube shaped box for packaging.

Explanation:

Consider the cylindrical shape box,

The ratio of volume to surface area of cylindrical shaped box will be as follows.

  $\begin{array}{c}{R}_{cylinder}=\frac{V_{cylinder}}{S{A}_{cylinder}}\\ =\frac{rh}{2{\left(r+h\right)}^{\prime }}\end{array}$

The ratio of volume to surface area of cube shaped box will be as follows.

  $\begin{array}{c}{R}_{cube}=\frac{V_{cube}}{S{A}_{cube}}\\ =\frac{s^3}{6s^2}\\ =\frac{s}{6}\end{array}$

If $r=h=s$ , then

  $\begin{array}{c}{R}_{cylinder}=\frac{r^2}{2{\left(r+r\right)}^{\prime }}\\ =\frac{r^2}{4^{\prime }r}\\ =\frac{r}{4^{\prime }}\end{array}$

  $R_{cube}=\frac{r}{6}$

It can be seen that $R_{cylinder}>R_{cube}$ .

Thus, cylindrical box is more efficient.

If now suppose $h=s=2r$ , then

  $\begin{array}{c}{R}_{cylinder}=\frac{r\left(2r\right)}{2{\left(r+2r\right)}^{\prime }}\\ =\frac{2r^2}{6^{\prime }r}\\ =\frac{1r}{3^{\prime }}\end{array}$

  $\begin{array}{c}{R}_{cube}=\frac{2r}{6}\\ =\frac{1r}{3}\end{array}$

It can be seen that $R_{cylinder}=R_{cube}$ .

Thus, both boxes are equally efficient.

So, the cereal’s box efficiency can be measured by comparing their ratio of volume to surface area and assuming different scenarios of radius, height, and side of cube.