Optimization problem

The volume of the cylindrical tin is \[V=\pi r^2 h,\] where $V$ is a given constant. Solving for $h$ we get \[h=\frac{V}{\pi r^2}.\] The cost of building the can is given by \[f=4c (\pi r^2)+c(2\pi rh)=4\pi c r^2+2\pi c r (\frac{V}{\pi r^2})=2c(2\pi r^2+\frac{V}{r})\] \[\Rightarrow f=2c(2\pi r^2+\frac{V}{r}).\]

Indeed $4c (\pi r^2)=2c \times 2 \times \pi r^2$ is the cost of building the ends and $c(2\pi rh)$ is the cost of building the sides. 

We want to minimze $f$. Hence compute \[f'(r)=2c (4\pi r-\frac{V}{r^2})=0 \Rightarrow 4\pi r^3=V\] \[ \Rightarrow r=(\frac{V}{4\pi})^{1/3}.\] Also \[h=\frac{V}{\pi r^2}=\frac{V}{\pi (\frac{V}{4\pi})^{2/3}}=\frac{V}{\pi \frac{V^{2/3}}{(4\pi)^{2/3}}}=\frac{4^{2/3}V^{1/3}}{\pi^{1/3}}=\frac{4V^{1/3}}{(4\pi)^{1/3}}\]
\[=4(\frac{V}{4\pi})^{1/3}=4r.\]

In other words, the cost will be minimized if \[r=(\frac{V}{4\pi})^{1/3}, h= 4(\frac{V}{2\pi})^{1/3},\] or when \[h=4r.\]