Volumes of aluminum cans

$\newenvironment {prompt}{}{} \newcommand {\accordion }[0]{} \newcommand {\ungraded }[0]{} \newcommand {\xmDefaultPreamble }[0]{xmPreamble.tex} \newcommand {\xmDefaultPrintstyle }[0]{xmPrintstyle.sty} \newcommand {\todo }[0]{} \newcommand {\oiint }[0]{{\large \bigcirc }\kern -1.56em\iint } \newcommand {\mooculus }[0]{\textsf {\textbf {MOOC}\textnormal {\textsf {ULUS}}}} \newcommand {\sagemath }[0]{\textsf {SageMath}} \newcommand {\surfaceColor }[0]{violet} \newcommand {\surfaceColorTwo }[0]{redyellow} \newcommand {\sliceColor }[0]{greenyellow} \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\sectionOutcomes }[0]{}$

Two young mathematicians discuss optimizing aluminum cans.

Check out this dialogue between two calculus students (based on a true story):

Devyn: Riley, have you ever noticed aluminum cans?
Riley: So very recyclable!
Devyn: I know! But I’ve also noticed that there are some that are short and fat, and others that are tall and skinny, and yet they can still have the same volume!
Riley: So very observant!
Devyn: This got me wondering, if we want to make a can with volume $V$, what shape of can uses the least aluminum?
Riley: Ah! This sounds like a job for calculus! The volume of a cylindrical can is given by
\[ V = \pi \cdot r^2 \cdot h \]
where $r$ is the radius of the cylinder and $h$ is the height of the cylinder. Also the surface area is given by \begin{align*} A &= \underbrace {\pi \cdot r^2}_{\text {bottom}} + \underbrace {2\cdot \pi \cdot r\cdot h}_{\text {sides}} + \underbrace {\pi \cdot r^2}_{\text {top}}\\ &= 2\cdot \pi \cdot r^2 + 2\cdot \pi \cdot r\cdot h. \end{align*}
Somehow we want to minimize the surface area, because that’s the amount of aluminum used, but we also want to keep the volume constant.
Devyn: Whoa, we have way too many letters here.
Riley: Yeah, somehow we need only one variable. Yikes. Too many letters.

Suppose we wish to construct an aluminum can with volume $V$ that uses the least amount of aluminum. In the context above, what do we want to minimize?

$A$ $V$ $h$ $r$

In the context above, what should be considered a constant?

$A$ $V$ $h$ $r$

As Devyn and Riley noticed, when we work out this type of problem, we need to reduce the problem to a single variable.

Consider $r$ to be the variable, and express $A$ as a function of $r$.

First, let’s eliminate the variable $h$. We know that $V$ is a constant and that

\[ V = \pi \cdot r^2 \cdot h \]

so $h=V/(\pi r^2)$. Substitute this expression for $h$ in the equation

\[ A = 2\cdot \pi \cdot r^2 + 2\cdot \pi \cdot r \cdot h. \]

\[ A(r) = 2\cdot \pi \cdot r^2 + 2 \cdot V/ \answer [given]{r} \]

Now consider $h$ to be the variable, and express $A$ as a function of $h$.

First, let’s eliminate the variable $r$. We know that $V$ is a constant and that

\[ V = \pi \cdot r^2 \cdot h \]

so $r=\sqrt {V/(\pi h)}$. Substitute this expression for $h$ in the equation

\[ A = 2\cdot \pi \cdot r^2 + 2\cdot \pi \cdot r \cdot h. \]

\[ A(h) = 2\cdot V/\answer [given]{h}+ 2 \cdot \sqrt {V\pi h} \]

Notice that we’ve reduced (one way or another) this function of two variables to a function of one variable. This process will be a key step in nearly every problem in this next section.

2025-01-06 19:07:15