Chapter 1.8, Problem 26E

a.

To determine

To find: the inequality.

Answer to Problem 26E

  $\begin{array}{l}7.998\le l\le 8.002\\ 4.248\le w\le 4.252\end{array}$

Explanation of Solution

Given information:

  $1\text{ mil}=\frac{1}{1000}\text{ inch}$

Possible dimensions = $8\text{ inch}\times \text{4}\frac{1}{4}\text{ inch}$

Accuracy = 2 mils

Calculation:

Let length and width be l and w respectively.

So, the required accuracy of the length is as follows:

  $\begin{array}{l}\left|l-8\right|\le \frac{2}{1000}\\ \left|l-8\right|\le 0.002\end{array}$

The above inequality can be rewritten by absolute value definition as follows:

  $\begin{array}{c}-0.002\le l-8\le 0.002\\ -0.002+8\le l-8+8\le 0.002+8\left[\text{Add 8}\right]\\ 7.998\le l\le 8.002\end{array}$

So, the required accuracy of the width is as follows:

  $\begin{array}{l}\left|w-4\frac{1}{4}\right|\le \frac{2}{1000}\\ \left|w-4.25\right|\le 0.002\end{array}$

The above inequality can be rewritten by absolute value definition as follows:

  $\begin{array}{c}-0.002\le w-4.25\le 0.002\\ -0.002+4.25\le w-4.25+4.25\le 0.002+4.25\left[\text{Add 4}\text{.25}\right]\\ 4.248\le w\le 4.252\end{array}$

b.

To determine

To graph: the region showing satisfactory dimensions for the part.

Answer to Problem 26E

For $7.998\le l\le 8.002$ :

  

For $4.248\le w\le 4.252$ :

  

Explanation of Solution

Given information:

  $\begin{array}{l}7.998\le l\le 8.002\\ 4.248\le w\le 4.252\end{array}$

Calculation:

The region showing satisfactory dimensions for the part $7.998\le l\le 8.002$ is as follows:

  

The region showing satisfactory dimensions for the part $4.248\le w\le 4.252$ is as follows: