Q. 22 - Precalculus Enhanced with Graphing Utilities

Question

Metro Media 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 \(500 per mile along the road and \)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. Give the domain.

(b) Compute the cost if $x=1$ mile.

(c) Compute the cost if $x=3$ 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?

Step-by-Step Solution

Answer

Part (a). The total cost C of installing the cable as a function of x is $C\left(x\right)=500x+700\sqrt{x^2-10x+29}$. The domain of C(x) is $[0,5]$.

Part (b). $C\left(1\right)=3630.495$dollars.

Part (c). $C\left(3\right)=3479.89$ dollars.

Part (d). As x increases from 0 to 5 , the value of C(x) decreases to a minimum and then increase.

Part (e). The values of x as $2.97$ results in least cost.

Part (a) Step 1. Given information and using Pythagorean theorem.

Consider the given statement as the house of customer 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.

As it is given that, if the installation cost is $500 per mile along the road and $700 per mile off the road.

The length of the cable along the road is x mile and cost per mile is $500, so the cost of installing the cable along the road is $500x. 

Now by Pythagorean Theorem,

$$\begin{gathered} d=\sqrt{2^2+{(5-x)}^2} \\ d=\sqrt{4+25+x^2-10x} \\ d=\sqrt{x^2-10x+19} \end{gathered}$$

Part (a) Step 2. Find the total cost of installation and domain of the function.

As the length of the cable off the road is $d=\sqrt{x^2-10x+29}$ and the cost of installation off the road is $700 per mile. The total cost C(x) of installing the cable as a function of x is, 

$C\left(x\right)=500x+700 \sqrt{x^2-10x+29}$

Now the domain of C(x) is the set of all values of x on which C(x) is defined. 

Now from the diagram, it can be seen that x could be 0 or x could be 5 or x could be any number between 0 and 5 . 

Therefore, the domain of C(x) is $[0,5]$.

Part (b) Step 1. Substitute x = 1 mile in C ( x ) = 500 x + 700 x 2 - 10 x + 29

This gives

$$\begin{gathered} C\left(1\right)=500+700\sqrt{1-10+29} \\ C\left(1\right)=500+700\sqrt{20} \\ C\left(1\right)=3630.495 \end{gathered}$$

Here the cost is $3630.495$ dollars.

Part (c) Step 1. Substitute x = 3 miles in C ( x ) = 500 x + 700 x 2 - 10 x + 29 .

This gives

$$\begin{gathered} C\left(3\right)=500\left(3\right)+700\sqrt{3^2-10\left(3\right)+29} \\ C\left(3\right)=1500+700\sqrt{8} \\ C\left(3\right)=3479.89 \end{gathered}$$

Here the cost is $3479.89$ dollars.

Part (d) Step 1. The graph of the function C ( x ) = 500 x + 700 x 2 - 10 x + 29 is as follows.

The graph can be traced for the values of x from 0 to 5.

As x increases from 0 to 5, the value of C(x) decreases to a minimum and the increase.

Part (e) Step 1. Find the local minima using graph from previous part.

From the graph, local minima exists at $C(2.97)\approx 3479.81$.

So least cost is approximately $3479.81$ dollars at $x=2.79$ miles.