Chapter 5.8, Problem 56PPS

(a)

To determine

To Draw: Graph of inequality

Explanation of Solution

Given:

  $\begin{array}{l}\text{Outer diameter of the pipe is to be}=10in\\ \text{Minimum cross section area of the pipe to be}=35f{t}^2\\ \text{Maximum cross section area of the pipe to be=}42f{t}^2\end{array}$

  

Formula Used:

  $\begin{array}{l}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ \text{cross section area }=\frac{\pi R^2}{4}-\frac{\pi r^2}{4}\end{array}$

Calculation:

  $\begin{array}{l}\text{Outer diameter of the pipe is to be}=10in\\ \text{Minimum cross section area of the pipe to be}=35f{t}^2\\ \text{Maximum cross section area of the pipe to be=}42f{t}^2\\ 35\le \text{cross section area}\le \text{42}\\ \Rightarrow \text{35}\le \frac{\pi }{4}\left(100-x^2\right)\le 42\\ LHS\\ \Rightarrow \frac{140}{\pi }\le \left(100-x^2\right)\\ \Rightarrow x^2\le 100-\frac{140}{\pi }\\ \Rightarrow x^2\le \frac{2200-980}{22}\\ \Rightarrow x\le \sqrt{\frac{1220}{22}}=7.446in\\ RHS\\ \left(100-x^2\right)\le \frac{168}{\pi }\\ \Rightarrow x^2\ge 100-\frac{168}{\pi }\\ \Rightarrow x\ge \sqrt{\frac{2200-1176}{22}}\\ \Rightarrow x\ge 6.821in\\ 7.446\le x\le 6.821\end{array}$

  

(b)

To determine

To Find: The thickness of the pipe.

Answer to Problem 56PPS

  $2.554in$ & $3.179in$

Explanation of Solution

Given:

  $\begin{array}{l}\text{Outer diameter of the pipe is to be}=10in\\ \text{Minimum cross section area of the pipe to be}=35f{t}^2\\ \text{Maximum cross section area of the pipe to be=}42f{t}^2\end{array}$

  

Formula Used:

  $\begin{array}{l}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ \text{cross section area }=\frac{\pi R^2}{4}-\frac{\pi r^2}{4}\end{array}$

Calculation:

  $\begin{array}{l}\text{Outer diameter of the pipe is to be}=10in\\ \text{Minimum cross section area of the pipe to be}=35f{t}^2\\ \text{Maximum cross section area of the pipe to be=}42f{t}^2\\ 35\le \text{cross section area}\le \text{42}\\ \Rightarrow \text{35}\le \frac{\pi }{4}\left(100-x^2\right)\le 42\\ LHS\\ \Rightarrow \frac{140}{\pi }\le \left(100-x^2\right)\\ \Rightarrow x^2\le 100-\frac{140}{\pi }\\ \Rightarrow x^2\le \frac{2200-980}{22}\\ \Rightarrow x\le \sqrt{\frac{1220}{22}}\le 7.446in\\ RHS\\ \left(100-x^2\right)\le \frac{168}{\pi }\\ \Rightarrow x^2\ge 100-\frac{168}{\pi }\\ \Rightarrow x\ge \sqrt{\frac{2200-1176}{22}}\\ \Rightarrow x\ge 6.821in\\ 7.446\ge x\ge 6.821\\ \text{Minimum thickness of the pipe}=10-7.446=2.554in\\ \text{Maximum thickness of the pipe}=10-6.821=3.179in\end{array}$

Conclusion:

Thus,

  $\begin{array}{l}\text{Minimum thickness of the pipe}=2.554in\\ \text{Maximum thickness of the pipe}=3.179in\end{array}$