Chapter 12.3, Problem 22PPS

To determine

To find: the minimum and maximum volumes of a fire hose in cubic feet.

Answer to Problem 22PPS

The maximum volume is approximately $\text{9}{\text{.81ft}}^3$ and the minimum volume is approximately $0.61{\text{ft}}^3$ .

Explanation of Solution

Given:

The diameters of fire hoses range from 1.5 in. to 6 in.

Thelength of the hoses is 50 ft.

Calculation:

First convert the inches of the diameter into feet. Use the fact that

  $\begin{array}{l}1\text{ft}=12\text{in}\text{.}\\ 12\text{in}\text{.}=1\text{ft}\\ 1\text{in}\text{.=}\frac{1}{12}\text{ft}\end{array}$

Convert 1.5 in. into feet.

  $\begin{array}{c}1\text{in}\text{.}=\frac{1}{12}\text{ft}\\ 1\times 1.5\text{in}\text{.}=\frac{1}{12}\times 1.5\text{ft} \left[\text{Multiply both sides by 1}\text{.5}\right]\\ 1.5\text{in}\text{.}=\frac{1.5}{12}\text{ft}\\ \text{=0}\text{.125ft}\end{array}$

Next convert 6 in. into feet.

  $\begin{array}{c}1\text{in}\text{.}=\frac{1}{12}\text{ft}\\ 1\times 6\text{in}\text{.}=\frac{1}{12}\times 6\text{ft} \left[\text{Multiply both sides by 6}\right]\\ \text{6in}\text{.}=\frac{6}{12}\text{ft}\\ \text{=0}\text{.5ft}\end{array}$

Now, find the volume of the hose having diameter 1.5 in. that is 0.125 ft and the length of the hose 50 ft.

The volume V of the cylinder is given by the formula

  $V=B\cdot h$

where $B=\pi r^2$ is the area of the circle and h is the height the cylinder.

Since radius is half the diameter of a circle, so radius of the circle is $\frac{0.125}{2}$ ft that is 0.0625 ft

The area B of the circle is given

  $\begin{array}{c}B=\pi r^2\\ =\pi {\left(0.0625\right)}^2\\ \approx 0.0039\pi \\ \approx 0.0039\left(3.14\right)\left[Since\pi \approx 3.14\right]\\ \approx 0.0122\end{array}$

Put $B\approx 0.0122$ and $h=50$ in the volume formula to find the volume.

Thus, the volume $V_1$ of the hose is

  $\begin{array}{c}{V}_1=B\cdot h\\ \approx \left(0.0122\right)50\\ =0.61\end{array}$

Therefore, the volume of the hose is $0.61{\text{ft}}^3$ . The minimum volume is approximately $0.61{\text{ft}}^3$ .

Now, find the volume of the hose having diameter 6 in. that is 0.5 ft and the length of the hose 50 ft.

The volume V of the cylinder is given by the formula

  $V=B\cdot h$

where $B=\pi r^2$ is the area of the circle and h is the height the cylinder.

Since radius is half the diameter of a circle, so radius of the circle is $\frac{0.5}{2}$ ft that is 0.25 ft.

The area B of the circle is given

  $\begin{array}{c}B=\pi r^2\\ =\pi {\left(0.25\right)}^2\\ \approx 0.0625\pi \\ \approx 0.0625\left(3.14\right)\left[Since\pi \approx 3.14\right]\\ \approx 0.19625\end{array}$

Put $B\approx 0.19625$ and $h=50$ in the volume formula to find the volume.

Thus, the volume $V_2$ of the hose is

  $\begin{array}{c}{V}_2=B\cdot h\\ \approx \left(0.19625\right)50\\ \approx \text{9}\text{.81}\end{array}$

Therefore, the volume of the hose is $\text{9}{\text{.81ft}}^3$ . The maximum volume is approximately $\text{9}{\text{.81ft}}^3$ .

Conclusion:

Therefore, the maximum volume is approximately $\text{9}{\text{.81ft}}^3$ and the minimum volume is approximately $0.61{\text{ft}}^3$ .