Question

The volume of a cylindrical tin can with a top and a bottom is $16\mathrm{\pi }$ cubic inches.

If a minimum amount of tin is to be used to construct the can, what must be the height of the can in inches $?$

• A. Height$=2\sqrt[3]{2}$ inches.
• B. Height$=2\sqrt[3]{4}$ inches.
• C. Height$=4$ inches.
• D. Height$=8$ inches

Answer 1

The correct option is C

Height$=4$ inches.

Step-1: Explanation for correct option:

It is given that the volume of the cylindrical tin can is$16\mathrm{\pi }$ cubic inches.

Let $\text{h}$ be the height and $\text{r}$ be the radius:

The formula of volume $\text{(V)}$ of cylindrical can is as follows:

$\mathbf{V}\mathbf{=}{\mathbf{\pi r}}^{\mathbf{2}}\mathbf{h}$

Since the volume is $16\mathrm{\pi }$, it follows that:

$$\begin{gathered} 16\mathrm{\pi }={\mathrm{\pi r}}^2\mathrm{h} \\ \Rightarrow \mathrm{h}=\frac{16}{{\mathrm{r}}^2} \end{gathered}$$

(first equation)

Let $\text{A}$ be the minimum surface area needed to use least amount of tin:

$\mathbf{\text{A}}\mathbf{=}\mathbf{2}{\mathbf{\pi r}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{nrh}$ (formula to find the surface area of cylindrical object)

(substitute the value of $\text{h}$ from the first equation in the above equation)

$\begin{array}{rcl}A& =& 2{\mathrm{\pi r}}^2+2\mathrm{\pi r}\left(\frac{16}{{\mathrm{r}}^2}\right)\\ & \Rightarrow & =2{\mathrm{\pi r}}^2+\frac{32\mathrm{\pi }}{\mathrm{r}}\end{array}$

Now, $\frac{dA}{dr}=4\mathrm{\pi r}-\frac{32\mathrm{\pi }}{{\mathrm{r}}^2}=0$ at max/min:

$\begin{array}{rcl}& \Rightarrow & 4\mathrm{\pi r}=\frac{32\mathrm{\pi }}{{\mathrm{r}}^2}\\ & \Rightarrow & 4{\mathrm{\pi r}}^3=32\mathrm{\pi }\\ & \Rightarrow & {\mathrm{r}}^3=8\\ \therefore \mathrm{r}& =& 2\end{array}$

Substitute the value of $\text{r}$ in the first equation:

$$\begin{gathered} h=\frac{16}{2^2} \\ =4 \end{gathered}$$

Hence, the height of cylindrical tin can is $\mathbf{\text{4 inches.}}$

Step-2: Explanation for incorrect option:

Since the height of can is $4$ inches, not equivalent to option $\text{(A)}$, so option $\text{(A)}$ is incorrect.

Since the height of can is $4$ inches, not equivalent to option $\text{(B)}$, so option $\text{(B)}$ is incorrect.

Since the height of can is $4$ inches, not equivalent to option $\text{(D)}$, so option $\text{(D)}$ is incorrect.

Hence, the only correct option is $\mathbf{\text{(C)}}$.