Chapter 7 Differential Calculus

Section 7.5 Applications

7.5.5 Example

Let us consider the example above in more detail. Obviously, the problem is to minimise the surface area of a cylindrical can with a given volume (base multiplied by height):

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{V=\pi r^{2}h=1\hspace{0.5em},}\end{array}}$
where $r$ is the radius of the base and $h$ is the height of the can. The surface area consists of the lid and the base (both with an area of $\pi r^2$) and the lateral surface (with an area of $2\pi rh$), which results in the equation $O=2\pi r^2+2\pi rh$ for the surface area of the can. The surface area of the can is a function of the radius $r$ and the height $h$. In contrast, a fixed volume (constraint) is assigned to the volume. Thus, it can be written:

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{O(r,h)=2\pi r^2+2\pi rh\hspace{0.5em}.}\end{array}}$
Due to the constraint $V=\pi r^{2}h=1$, this problem that initially involves two variables ($r$ and $h$) can be reduced to a problem that only involves one variable. Solving the constraint for the height of the can results in:

${\displaystyle \begin{array}{ccc}\multicolumn{1}{c}{\pi r^{2}h}& =\hfill & 1\hfill \\ \multicolumn{1}{c}{\iff \mathrm{ }h}& =\hfill & \frac{1}{\pi r^2}\hspace{0.5em}.\hfill \end{array}}$
Substituting this formula into the function $O(r,h)$ results in a function that only depends on one variable. This function is also called $O$ for simplicity:

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{O(r,h)=2\pi r^2+2\pi rh=2\pi r^2+2\pi r\frac{1}{\pi r^2}=2(\pi r^2+\frac{1}{r})=O\left(r\right)\hspace{0.5em}.}\end{array}}$
After this manipulation, the problem of finding the cans minimal surface area can be solved analogously to normal extremal value problems of functions. Thus, we take the first derivative of the function $O$ with respect to the variable $r$ and set this derivative equal to zero:

${\displaystyle \begin{array}{ccc}\multicolumn{1}{c}{O\text{'}\left(r\right)=2(2\pi r-\frac{1}{r^2})}& =\hfill & 0\hfill \\ \multicolumn{1}{c}{\iff \mathrm{ }2\pi r}& =\hfill & \frac{1}{r^2}\hfill \\ \multicolumn{1}{c}{\iff \mathrm{ }2\pi r^3}& =\hfill & 1\hfill \\ \multicolumn{1}{c}{\iff \mathrm{ }{r}^3}& =\hfill & \frac{1}{2\pi }\hfill \\ \multicolumn{1}{c}{\iff \mathrm{ }r}& =\hfill & \sqrt[3]{\frac{1}{2\pi }}\hspace{0.5em}.\hfill \end{array}}$
The last equivalent transformation used the fact that the radius $r$ cannot take negative values. Substituting this result into the second derivative of $O$ shows whether a minimum was actually found ($O\text{'}\text{'}(r)=4\pi +4/r^3$):

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{O\text{'}\text{'}\left(\sqrt[3]{\frac{1}{2\pi }}\right)=4\pi +\frac{4}{{\left(\sqrt[3]{\frac{1}{2\pi }}\right)}^3}=12\pi >0\hspace{0.5em}.}\end{array}}$
For the radius $r=\sqrt[3]{\frac{1}{2\pi }}$, the surface area of the cylindrical can with the given volume $V=1$ is a minimum. The corresponding height of the can is $h=\frac{1}{\pi {\left(\sqrt[3]{\frac{1}{2\pi }}\right)}^2}=\sqrt[3]{\frac{4}{\pi }}$. If a can of these dimensions is manufactured, the material usage for the given volume is minimised.