Question

For a closed cylinder with radius $rcm$ and height $hcm$, find the dimensions giving the minimum surface area, given that the volume is $36c{m}^3$.

Answer 1

Finding the dimensions of the given closed cylinder:

Step-1: Finding the surface area and the volume of a closed cylinder

We know that the surface area and the volume of a closed cylinder with radius $runit$ and height $hunit$ are respectively: $2\pi rh+2\pi r^{2}uni{t}^2$ and $\pi r^{2}huni{t}^3$

Now, given that the volume is $36c{m}^3$. So, we get:

$$\begin{gathered} \pi r^{2}h=36 \\ \Rightarrow h=\frac{36}{\pi r^2} \end{gathered}$$

Step-2: Constructing the function to be minimized

The surface area of the given cylinder is

$$\begin{gathered} S=2\pi rh+2\pi r^2 \\ =2\pi r\times \frac{36}{\pi r^2}+2\pi r^2 \\ =\frac{72}{r}+2\pi r^2 \end{gathered}$$

Consider $f\left(r\right)=\frac{72}{r}+2\pi r^2$.

Step-3: Finding the critical points of $f\left(r\right)$

Differentiating $f\left(r\right)$ with respect to $r$, we get: $f\text{'}\left(r\right)=-\frac{72}{r^2}+4\pi r$.

We know that the critical points of $f\left(r\right)$ will be the solutions of $f\text{'}\left(r\right)=0$.

Now,

$$\begin{gathered} f\text{'}\left(r\right)=0 \\ \Rightarrow -\frac{72}{r^2}+4\pi r=0 \\ \Rightarrow \frac{-72+4\pi r^3}{r^2}=0 \\ \Rightarrow -72+4\pi r^3=0 \\ \Rightarrow r^3=\frac{72}{4\pi } \\ \Rightarrow r^3=\frac{72\times 7}{4\times 22} \\ \therefore r=\sqrt[3]{\frac{63}{11}} \end{gathered}$$

So, the critical point of $f\left(r\right)$ is $r=\sqrt[3]{\frac{63}{11}}$.

Step-4: Checking whether $r=\sqrt[3]{\frac{63}{11}}$ is a point of minimum of $f\left(r\right)$

Differentiating $f\text{'}\left(r\right)$, with respect to $r$, we get: $f\text{'}\text{'}\left(r\right)=\frac{144}{r^3}+4\pi$.

Now, $r=\sqrt[3]{\frac{63}{11}}$ will be a point of minimum of $f\left(r\right)$ if $f\text{'}\text{'}\left(\sqrt[3]{\frac{63}{11}}\right)>0$.

Now,

$$\begin{gathered} f\text{'}\text{'}\left(\sqrt[3]{\frac{63}{11}}\right) \\ =\frac{144}{{\displaystyle \frac{63}{11}}}+4\times \frac{22}{7} \\ =\frac{2376}{63}>0 \end{gathered}$$

Hence, $r=\sqrt[3]{\frac{63}{11}}$ is a point of minimum of $f\left(r\right)$.

Step-5: Finding the radius and height of the cylinder

From Step-4, we see that: $r=\sqrt[3]{\frac{63}{11}}=1.79cm$ and hence

$$\begin{gathered} h=\frac{36}{\pi r^2} \\ =\frac{36}{{\displaystyle \frac{22}{7}}\times {\left({\displaystyle \frac{63}{11}}\right)}^{{\displaystyle \frac{2}{3}}}} \\ =3.58cm \end{gathered}$$

Therefore, the dimensions of the given cylinder are : $r=1.79cm$ and $h=3.58cm$.