Chapter B.5, Problem 4E

To determine

Whether the viewing rectangle results in a square screen for

$\begin{array}{lll}X\text{min}=-16\hfill & X\text{max}\text{=}16\hfill & X\text{scl}\text{=}8\hfill \\ Y\text{min}\text{=}-10\hfill & Y\text{max}\text{=}10\hfill & Y\text{scl}\text{=}5\hfill \end{array}$.

Solution:

Yes. The viewing rectangle results in a square screen.

Given information:

$\begin{array}{lll}X\text{min}=-16\hfill & X\text{max}\text{=}16\hfill & X\text{scl}\text{=}8\hfill \\ Y\text{min}\text{=}-10\hfill & Y\text{max}\text{=}10\hfill & Y\text{scl}\text{=}5\hfill \end{array}$

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.

$\begin{array}{l}\begin{array}{cc}X\text{min}=-16& Y\text{min}\text{=}-10\end{array}\\ \begin{array}{cc}X\text{max}\text{=}16& Y\text{max}\text{=}10\end{array}\end{array}$

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

$\Rightarrow \frac{X\text{max}-X\text{min}}{Y\text{max}-Y\text{min}}=\frac{16-\left(-16\right)}{10-\left(-10\right)}=\frac{32}{20}=\frac{8}{5}$

Therefore, the viewing rectangle results in a square screen.