Chapter 16, Problem 1P

Design the optimal cylindrical container (Fig. P16.1) that is open at one end and has walls of negligible thickness. The container is to hold $0.5{\text{ m}}^{\text{3}}$. Design it so that the areas of its bottom and sides are minimized.

FIGURE P16.1

To determine

The design of the optimal cylindrical container to hold $0.5{\text{ m}}^3$, provided area of its bottoms and sides are minimized.

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Answer to Problem 1P

Solution: The optimal design of the optimal cylinder has radius and height same and its values is $0.542\text{ m}$.

Explanation of Solution

Given Information:

The volume of the cylinder is $0.5{\text{ m}}^3$.

Formula used:

Write the expression of surface area of a cylindrical container.

$A=\pi r^2+2\pi rh$

Here, $r$ is the radius of the cylindrical container, and $h$ is the height of the cylindrical container.

Write the expression of volume of a cylindrical container.

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

Calculation:

Recall the expression of volume of a cylindrical container.

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

The container holds $0.5{\text{m}}^{\text{3}}$, therefore,

$\begin{array}{c}V=0.5\\ \pi r^{2}h=0.5\end{array}$.

Rearrange the expression of volume for height $h$.

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

Recall the expression of surface area of a cylindrical container.

$A=\pi r^2+2\pi rh$

Substitute $\frac{V}{\pi r^2}$ for $h$.

$\begin{array}{c}A=\pi r^2+2\pi r\left(\frac{V}{\pi r^2}\right)\\ =\pi r^2+\frac{2V}{r}\end{array}$

To find the conditions to minimize the area solve $\frac{dA}{dr}=0$.

$\begin{array}{c}\frac{dA}{dr}=0\\ \frac{d}{dr}\left(\pi r^2+\frac{2V}{r}\right)=0\\ 2\pi r+\frac{-2V}{r^2}=0\\ 2\pi r-\frac{2V}{r^2}=0\end{array}$

Solve further,

$\begin{array}{c}2\pi r-\frac{2V}{r^2}=0\\ r^3=\frac{V}{\pi }\\ r=\sqrt[3]{\frac{V}{\pi }}\end{array}$

Volume of the container remains constant therefore, substitute the value of $r$ in the expression of $h$ obtained from volume expression:

$\begin{array}{c}h=\frac{V}{\pi {\left(\frac{V}{\pi }\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\\ =\sqrt[3]{\frac{V}{\pi }}\end{array}$

Therefore, the value of height and radius are same to obtain to minimize the surface areas.

$\begin{array}{c}h=r\\ =\sqrt[3]{\frac{V}{\pi }}\end{array}$

Substitute $0.5{\text{ m}}^3$ for $V$.

$\begin{array}{c}h=\sqrt[3]{\frac{0.5}{\pi }}\\ =\text{0}\text{.542}\text{m}\end{array}$

Therefore, the value of the radius and the height to obtain the maximum surface area is $0.542\text{ m}$.