Chapter B.5, Problem 5E

To determine

Whether the viewing rectangle results in a square screen for

$\begin{array}{lll}X\text{min}=-6\hfill & X\text{max}\text{=}6\hfill & X\text{scl}\text{=}1\hfill \\ Y\text{min}\text{=}-2\hfill & Y\text{max}\text{=}2\hfill & Y\text{scl}\text{=}0.5\hfill \end{array}$.

Solution:

No. The viewing rectangle does notresult in a square screen.

Given information:

$\begin{array}{lll}X\text{min}=-6\hfill & X\text{max}\text{=}6\hfill & X\text{scl}\text{=}1\hfill \\ Y\text{min}\text{=}-2\hfill & Y\text{max}\text{=}2\hfill & Y\text{scl}\text{=}0.5\hfill \end{array}$

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.

$\begin{array}{l}\begin{array}{cc}X\text{min}=-6& Y\text{min}\text{=}-2\end{array}\\ \begin{array}{cc}X\text{max}\text{=}6& Y\text{max}\text{=}2\end{array}\end{array}$

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

$\Rightarrow \frac{X\text{max}-X\text{min}}{Y\text{max}-Y\text{min}}=\frac{6-\left(-6\right)}{2-\left(-2\right)}=\frac{12}{4}=\frac{3}{1}$

Therefore, the viewing rectangle does not result in a square screen.