Chapter 2.6, Problem 22AYU

22. Installing Cable TV MetroMedia Cable is asked to provide service to a customer whose house is located 2 miles from the road along which the cable is buried. The nearest connection box for the cable is located 5 miles down the road. See the figure.

(a) If the installation cost is S500 per mile along the road and $\$700$ per mile off the road, build a model that expresses the total cost C of installation as a function of the distance x (in miles) from the connection box to the point where the cable installation turns off the road. Find the domain of $C=C\left(x\right)$.

(b) Compute the cost if $x\text{ = 1}$ mile.

(c) Compute the cost if $x\text{ = 3}$ miles.

(d) Graph the function $C=C\left(x\right)$. Use TRACE to see how the cost C varies as x changes from 0 to 5.

(e) What value of x results in the least cost?

To determine

To find:

a. To model that expresses the total cost $C$ of installation as a function of distance $x$. To find the domain of $C=C(x)$.

Answer to Problem 22AYU

Solution:

a. $C\left(x\right)=500x+700\times \sqrt{x^2-10x+29}\text{Domain of} C\left(x\right)= \left\{x|0<x<5\right\}$.

Explanation of Solution

Given:

1. The house is 2 miles from the road.

2. The nearest connection box is 5 miles down the road.

3. Cost of installation is $\$500$ per mile along the road and $\$700$ per off the road.

Calculation:

The can be expressed as the figure below.

Let $x$ be the distance along the road.

Let $d$ be the distance off the road.

a. The total cost of installation $C=500x+700d$

From the figure, $d=\sqrt{2^2+{\left(5-x\right)}^2}$.

The Cost of installation $C\left(x\right)=500x+700 \times \sqrt{2^2+{\left(5-x\right)}^2}$.

$C\left(x\right)=500x+700 \times \sqrt{4+25+x^2-10x}$

$=500x+700\times \sqrt{x^2-10x+29}$

$\text{Domain of} C\left(x\right)= \left\{x|0\le x\le 5\right\}$

To determine

To find:

b. To compute the cost if $x=1$ mile.

Answer to Problem 22AYU

Solution:

b. $C\left(1\right)=\$3630.5$.

Explanation of Solution

Given:

1. The house is 2 miles from the road.

2. The nearest connection box is 5 miles down the road.

3. Cost of installation is $\$500$ per mile along the road and $\$700$ per off the road.

Calculation:

The can be expressed as the figure below.

Let $x$ be the distance along the road.

Let $d$ be the distance off the road.

b. For $x=1$

$C\left(1\right)=500+700\times \sqrt{1-10+29}$

$=\$3630.5$

To determine

To find:

c. To compute the cost if $x=3$ miles.

Answer to Problem 22AYU

Solution:

c. $C\left(3\right)=\$3479.9$

Explanation of Solution

Given:

1. The house is 2 miles from the road.

2. The nearest connection box is 5 miles down the road.

3. Cost of installation is $\$500$ per mile along the road and $\$700$ per off the road.

Calculation:

The can be expressed as the figure below.

Let $x$ be the distance along the road.

Let $d$ be the distance off the road.

c. For $x=3$.

$C\left(3\right)=500\times 3+700\times \sqrt{9-30+29}$

$=\$3479.9$

To determine

To find:

d. To graph the function $C=C(x)$.

Answer to Problem 22AYU

Explanation of Solution

Given:

1. The house is 2 miles from the road.

2. The nearest connection box is 5 miles down the road.

3. Cost of installation is $\$500$ per mile along the road and $\$700$ per off the road.

Calculation:

The can be expressed as the figure below.

Let $x$ be the distance along the road.

Let $d$ be the distance off the road.

d. To graph $C\left(x\right)$:

Sl No	$x$ In miles	$C\left(3\right)$In $\$$
1.	0	$3769.6$
2	1	$3630.5$
3	2	$3523.9$
4	3	$3479.9$
5	4	$3565.2$
6	5	$3900.0$

To determine

To find:

e. To find the value of $x$ in which the cost is least..

Answer to Problem 22AYU

Solution:

e. $\text{The value of}C(x)\text{is least when}x=2.96\text{miles}$.

Explanation of Solution

Given:

1. The house is 2 miles from the road.

2. The nearest connection box is 5 miles down the road.

3. Cost of installation is $\$500$ per mile along the road and $\$700$ per off the road.

Calculation:

The can be expressed as the figure below.

Let $x$ be the distance along the road.

Let $d$ be the distance off the road.

e. The value of $C\left(x\right)$ is least when $x=2.96$ miles.