Chapter B.5, Problem 1E

To determine

Whether the viewing rectangle results in a square screen for

$\begin{array}{lll}X\text{min}=-8\hfill & X\text{max}\text{=}\text{8}\hfill & X\text{scl}\text{=}\text{2}\hfill \\ Y\text{min}\text{=}-5\hfill & Y\text{max}\text{=}\text{5}\hfill & Y\text{scl}\text{=}\text{1}\hfill \end{array}$.

Solution:

Yes. The viewing rectangle results in a square screen.

Given information:

$\begin{array}{lll}X\text{min}=-8\hfill & X\text{max}\text{=}\text{8}\hfill & X\text{scl}\text{=}\text{2}\hfill \\ Y\text{min}\text{=}-5\hfill & Y\text{max}\text{=}\text{5}\hfill & Y\text{scl}\text{=}\text{1}\hfill \end{array}$

Explanation:

The selections for $X\text{min},X\text{max},Y\text{min},\text{and}Y\text{max}$ must be adjusted so that the ratio of $x\text{to}y$ must be $8:5$, in order to acquire the square screen.

$\begin{array}{l}\begin{array}{cc}X\text{min}=-8& Y\text{min}\text{=}-5\end{array}\\ \begin{array}{cc}X\text{max}\text{=}\text{8}& Y\text{max}\text{=}\text{5}\end{array}\end{array}$

Then, the ratio of $x\text{to}y$ is:

$\Rightarrow \frac{X\text{max}-X\text{min}}{Y\text{max}-Y\text{min}}=\frac{8-\left(-8\right)}{5-\left(-5\right)}=\frac{16}{10}=\frac{8}{5}$

Therefore, the viewing rectangle results in a square screen.