Chapter 8.6, Problem 51PS

Solution

(a)

To determine : The inequalities for circular regions around the hotel and cafe in which the person can get wireless Internet access, taking the location of the person as the origin.

The inequalities for circular regions around the hotel and cafe in which the person can get wireless Internet access are ${\left(x-100\right)}^2+{\left(y+60\right)}^2\le {150}^2$ and ${\left(x+80\right)}^2+{\left(y+70\right)}^2\le {100}^2$ respectively.

Given information :

A person is in a park surfing the Internet on a wireless connection. A hotel's wireless transmitter is located $100$ yards east and $60$ yards south of that person. It has a range of $150$ yards. A cafe's transmitter is located $80$ yards west and $70$ yards south of that person. It has a range of $100$ yards.

Explanation :

The inequality used to describe the region inside a circle is ${\left(x-h\right)}^2+{\left(y-k\right)}^2\le r^2$ .

The range of the hotel’s wireless transmitter is $150$ yards, so the radius of the circular region around the hotel will be maximum of $150$ yards.

The hotel's wireless transmitter is located $100$ yards east and $60$ yards south of that person.

So,

  $\begin{array}{l}h=100\\ k=-60\\ r=150\end{array}$

Thus, the inequality for circular regions around the hotel in which the person can get wireless Internet access will be, ${\left(x-100\right)}^2+{\left(y+60\right)}^2\le {150}^2$ .

The range of the cafe’s wireless transmitter is $100$ yards, so the radius of the circular region around the cafe will be maximum of $100$ yards.

The cafe's transmitter is located $80$ yards west and $70$ yards south of that person.

So,

  $\begin{array}{l}h=-80\\ k=-70\\ r=100\end{array}$

Thus, the inequality for circular regions around the cafe in which the person can get wireless Internet access will be, ${\left(x+80\right)}^2+{\left(y+70\right)}^2\le {100}^2$ .

Therefore, inequalities for circular regions around the hotel and cafe in which the person can get wireless Internet access are ${\left(x-100\right)}^2+{\left(y+60\right)}^2\le {150}^2$ and ${\left(x+80\right)}^2+{\left(y+70\right)}^2\le {100}^2$ respectively.

(b)

To graph : The inequalities for circular regions around the hotel and cafe in which the person can get wireless Internet access, taking the location of the person as the origin.

Given information :

A person is in a park surfing the Internet on a wireless connection. A hotel's wireless transmitter is located $100$ yards east and $60$ yards south of that person. It has a range of $150$ yards. A cafe's transmitter is located $80$ yards west and $70$ yards south of that person. It has a range of $100$ yards.

Graph :

The inequalities for circular regions around the hotel and cafe in which the person can get wireless Internet access are ${\left(x-100\right)}^2+{\left(y+60\right)}^2\le {150}^2$ and ${\left(x+80\right)}^2+{\left(y+70\right)}^2\le {100}^2$ respectively.

Graph the inequalities.

  

Point A represents the place where the person is standing.

The distance of the person at $\left(0,0\right)$ from the location of hotel’s transmitter at $\left(100,-60\right)$ is,

  $\begin{array}{c}\sqrt{{\left(100-0\right)}^2+{\left(-60-0\right)}^2}\\ =\sqrt{10000+3600}\\ =\sqrt{13600}\\ \approx 117\text{ yards}\end{array}$

The distance of the person at $\left(0,0\right)$ from the location of cafe’s transmitter at $\left(-80,-70\right)$ is,

  $\begin{array}{c}\sqrt{{\left(-80-0\right)}^2+{\left(-70-0\right)}^2}\\ =\sqrt{6400+4900}\\ =\sqrt{11300}\\ \approx 106\text{ yards}\end{array}$

Interpretation :

The person is at a distance of $117\text{ yards}$ from the hotel’s transmitter and at a distance of $106\text{ yards}$ from the café’s transmitter, so he is only in one region.

(c)

To explain : The way to determine the overlapping of regions.

To find that the regions overlap or not compare the sum of radius of the circular region in which the person can get wireless Internet access with the distance from the hotel’s transmitter to the cafe’s transmitter.

Given information :

A person is in a park surfing the Internet on a wireless connection. A hotel's wireless transmitter is located $100$ yards east and $60$ yards south of that person. It has a range of $150$ yards. A cafe's transmitter is located $80$ yards west and $70$ yards south of that person. It has a range of $100$ yards.

Explanation :

Find the distance from the hotel’s transmitter to the cafe’s transmitter.

  $\begin{array}{c}\sqrt{{\left(-80-100\right)}^2+{\left(-70+60\right)}^2}\\ =\sqrt{32400+100}\\ =\sqrt{32500}\\ \approx 180\text{ yards}\end{array}$

Compare it with the sum of the radius of the circular region in which the person can get wireless Internet access.

The distance from the hotel’s transmitter to the cafe’s transmitter is $\le 250$ yards.

Therefore, the regions overlap.