Find the dimensions of a cylindrical can with a volume of \(206in^3\) that minimizes the surface area.
The “#10 can”is a standard sized can used by the restaurant industry
that holds about \(206in^3\) with a
diameter of \(6\frac{2}{16}in\)
and height of \(7in\). Does it seem
these dimensions were chosen with minimizaon in mind?
Surface area, A of a cylinder; \[
A = 2\pi r^2 +2\pi rh
\] Volume, V of a cylinder; \[
V = \pi r^2h
\] Given V = \(206in^3\), solve
for h; \[
\begin{aligned}
206 &= \pi r^2h \\
h &= \frac{206}{\pi r^2} \approx \frac{65.57184}{r^2}
\end{aligned}
\] Substitue h into surface area; \[
\begin{aligned}
A(r) &= 2\pi r^2 + 2\pi r(\frac{65.57184}{r^2}) \\
&= 2\pi r^2 + \frac{412}{r}
\end{aligned}
\] To minimize the surface area, differentiate A, and set to
zero; \[
\begin{aligned}
A'(r) = 4\pi r - \frac{412}{r^2} &= 0 \\
4\pi r &= \frac{412}{r^2} \\
r^3 &= \frac{412}{4 \pi} \\
r &= 3.200583 \approx 3.2
\end{aligned}
\] To check if r corresponds to a minimum, check if \(A''(r) > 0\) \[
\begin{aligned}
A''(r) &= 4\pi + 2\frac{412}{r^3} \\
&= 4\pi + \frac{824}{r^3}
\end{aligned}
\] Since r is always positive then \(A''(r)\) is always positive. Thus,
r = 3.2 is corresponds to a minimum
Now, solve for h; \[
\begin{aligned}
h &= \frac{65.57184}{r^2} \\
&= \frac{65.57184}{3.2^2} \\
&= 6.401167 \approx 6.4
\end{aligned}
\] The height and radius that would correspond to a minimum
surface area is \(height = 6.4in\) and
\(radius = 3.2in\)
From this, the dimensions of a standard can are not chosen with minimization in mind.