Q. 51 - Precalculus Enhanced with Graphing Utilities

Question

Developing a Linear Model from Data The following data represent the price p and quantity demanded per day q of 24" LCD monitors. 

Price p,(in dollars)	Quantity Demanded, q
150	100
200	80
250	60
300	40

(a) Plot the ordered pairs (p, q) in a Cartesian plane. 

(b) Show that quantity demanded q is a linear function of the price p. 

(c) Determine the linear function that describes the relation between p and q. 

(d) What is the implied domain of the linear function? 

(e) Graph the linear function in the Cartesian plane drawn in part (a). 

(f) Interpret the slope. (g) Interpret the values of the intercepts. 

Step-by-Step Solution

Answer

Part (a) The graph is 

Part (b)  The function is linear.

Part (c) The function is $q\left(p\right)=-\frac{2}{5}p+160$.

Part (d) The domain of the function is $\left\{p|0\le p\le 400\right\}$.$\left\{p|0\le p\le 400\right\}$

Part (e) The graph is 

Part (f) There is decrease in quantity demanded by $\frac{2}{5}$ per dollar.

Part (g)

• This represent if price of monitor is 0 the quantity demanded is 160.
• This represent if price of monitor is 400 the quantity demanded is 0.

Part (a) Step 1. Given information

The table represents, price p and quantity demanded per day q of 24" LCD monitors.

Price, p (in dollars)	Quantity Demanded, q
150	100
200	80
250	60
300	40

Part (a) Step 2. Explanation.

Plot the ordered pairs (p, q) in a Cartesian plane.

Part (b) Step 1. Explanation.

Show that quantity demanded q is a linear function of the price p. 

Find average rate of change.

Consider two points $\left(150,100\right),\left(200,80\right)$.

$$\begin{gathered} \frac{∆y}{∆x}=\frac{80-100}{200-150} \\ =-\frac{20}{50} \\ =-\frac{2}{5} \end{gathered}$$

Consider two points $\left(200,80\right),\left(250,60\right)$.

$$\begin{gathered} \frac{∆y}{∆x}=\frac{60-80}{250-200} \\ =-\frac{20}{50} \\ =-\frac{2}{5} \end{gathered}$$

Consider two points $\left(250,60\right),\left(300,40\right)$.

$$\begin{gathered} \frac{∆y}{∆x}=\frac{40-60}{300-250} \\ =-\frac{20}{50} \\ =-\frac{2}{5} \end{gathered}$$

Here $\frac{∆y}{∆x}$ is constant. So the function is linear.

Part (c) Step 1. Explanation.

Write the linear function that describes the relation between p and q. 

The standard form of linear function is $y=mx+b$. Where m is slope and b is y-intercept.

We have $\frac{∆y}{∆x}=-\frac{2}{5}$. This is slope of the function.

So the function becomes $q\left(p\right)=-\frac{2}{5}p+b$.

To find b, take any arbitrary point $\left(150,100\right)$.

$$\begin{gathered} 100=-\frac{2}{5}\left(150\right)+b \\ 100=-60+b \\ 160=b \end{gathered}$$

 The linear function that describes the relation between p and q is $q\left(p\right)=-\frac{2}{5}p+160$

 

Part (d) Step 1. Explanation.

Find the implied domain of the linear function.

The price p of LCD monitor is greater than equal to 0. 

So if $p\ge 0\Rightarrow q\left(p\right)\ge 0.$

$$\begin{gathered} q\left(p\right)\ge 0 \\ -\frac{2}{5}p+160\ge 0 \\ -\frac{2}{5}p\ge -160 \\ p\ge 400 \end{gathered}$$

The implied domain of the linear function is $\left\{p|0\le p\le 400\right\}$

Part (e) Step 1. Explanation.

The graph of linear function is as follows,

Part (f) Step 1. Explanation.

Interpret the slope. 

In the function $q\left(p\right)=-\frac{2}{5}p+160$, slope is $-\frac{2}{5}$.

 As slope is negative. the function is decreasing.

Here, There is decrease in quantity demanded by $\frac{2}{5}$per dollar.

Part (g) Step 1. Explanation.

Interpret the values of the intercepts. 

In the function $q\left(p\right)=-\frac{2}{5}p+160$ y-intercept is 160.

• This represent if price of monitor is 0 the quantity demanded is 160.

In the function when $q\left(p\right)=0$ then $p=400$. This is x-intercept.

• This represent if price of monitor is 400 the quantity demanded is 0.