Chapter 0.2, Problem 56E

a.

To determine

To Write: The function $C\left(x\right)$ which gives the cost of laying the cable in terms of the distance $x$ .

Answer to Problem 56E

  $C\left(x\right)=180\sqrt{{800}^2+x^2}+100\left(10,560-x\right)$

Explanation of Solution

Given:

Width of the river $=800\text{ft}$

Cost of laying the cable in the river $=180\$$ per foot

Cost of laying the cable in the river $=100\$$ per foot.

  

Calculation:

Given that,

In the opposite direction, the cable goes from the plant to point Q, which is $x$ feet away from point $P$ .

In general, $1\text{mile}=5280\text{ ft}$ , So

  $2\text{mi}=10,560\text{ft}$

As a result, the total cable length inside the river will be:

  $\sqrt{{800}^2+x^2}\text{ft}$

The Cable total length on the land will be:

  $\left(10,560-x\right)\text{ft}$

Thus, the cost to lay the cable in terms of $x$ is as follows:

  $C\left(x\right)=180\sqrt{{800}^2+x^2}+100\left(10,560-x\right)$

b.

To determine

To Generate: The table of values to determine if point $Q$ is less than $2000\text{ ft}$ or more than $2000\text{ ft}$ away from point $P$ .

Answer to Problem 56E

From point $P$ , the least expensive location appears to be less than $2000\text{ ft}$ away.

Explanation of Solution

Given:

Width of the river $=800\text{ft}$

Cost of laying the cable in the river $=180\$$ per foot

Cost of laying the cable in the river $=100\$$ per foot.

  

Calculation:

From part $\left(a\right)$ , the function is $C\left(x\right)=180\sqrt{{800}^2+x^2}+100\left(10,560-x\right)$

Substitute values to $x$ values to find if point $Q$ is less than $2000\text{ ft}$ .

Substitute $x=0$ in $C\left(x\right)=180\sqrt{{800}^2+x^2}+100\left(10,560-x\right)$ :

  $\begin{array}{c}C\left(0\right)=180\sqrt{{800}^2+0^2}+100\left(10,560-0\right)\\ =144000+100(10560-0)\\ =1200000\end{array}$

Substitute $x=500$ in $C\left(x\right)=180\sqrt{{800}^2+x^2}+100\left(10,560-x\right)$ :

  $\begin{array}{c}C\left(500\right)=180\sqrt{{800}^2+x^2}+100\left(10,560-500\right)\\ =180\sqrt{890000}+100(10560-500)\\ =180\sqrt{{100}^2\cdot 89}+100\cdot 10060\\ =180(100\sqrt{89})+1006000\\ =18000\sqrt{89}+1006000\\ =1175811.6603\\ \approx 1175812\end{array}$

Substitute $x=1000$ in $C\left(x\right)=180\sqrt{{800}^2+x^2}+100\left(10,560-x\right)$ :

  $\begin{array}{c}C\left(1000\right)=180\sqrt{{800}^2+{1000}^2}+100\left(10,560-1000\right)\\ =180\sqrt{640000+1000000}+100\left(956000\right)\\ =180\sqrt{1640000}+956000\\ =180\sqrt{{200}^2\cdot 41}+956000\\ =180\left(200\sqrt{41}\right)+956000\\ =36000\sqrt{41}+956000\\ \text{=1186512}\text{.47}\end{array}$

Substitute $x=1500$ in $C\left(x\right)=180\sqrt{{800}^2+x^2}+100\left(10,560-x\right)$ :

  $\begin{array}{c}C\left(1500\right)=180\sqrt{{800}^2+{1500}^2}+100\left(10,560-1500\right)\\ =180\sqrt{640000+2250000}+100\times 9060\\ =180\sqrt{2890000}+906000\\ =180\sqrt{{1700}^2}+906000\\ =180\times 1700+906000\\ =\text{306000}+906000\\ =1212000\end{array}$

Substitute $x=2000$ in $C\left(x\right)=180\sqrt{{800}^2+x^2}+100\left(10,560-x\right)$ :

  $\begin{array}{c}C\left(2000\right)=180\sqrt{{800}^2+{2000}^2}+100\left(10,560-2000\right)\\ =180\sqrt{640000+4000000}+100\times 8560\\ =180\sqrt{46400000}+856000\\ =180\sqrt{{400}^2\cdot 29}+856000\\ =180\left(400\sqrt{29}\right)+856000\\ =72000\sqrt{29}+856000\\ =1243731.86\\ \approx 1243732\end{array}$

Substitute $x=2500$ in $C\left(x\right)=180\sqrt{{800}^2+x^2}+100\left(10,560-x\right)$ :

  $\begin{array}{c}C\left(2500\right)=180\sqrt{{800}^2+{2500}^2}+100\left(10,560-2500\right)\\ =180\sqrt{640000+6250000}+100\left(8060\right)\\ =180\sqrt{6890000}+806000\\ =180\sqrt{{100}^2\cdot 689}+806000\\ =180\times 100\sqrt{689}+806000\\ =18000\sqrt{689}+806000\\ =1278478.57\\ \approx 1278479\end{array}$

Substitute $x=3000$ in $C\left(x\right)=180\sqrt{{800}^2+x^2}+100\left(10,560-x\right)$ :

  $\begin{array}{c}C\left(3000\right)=180\sqrt{{800}^2+{3000}^2}+100\left(10,560-3000\right)\\ =180\sqrt{640000+9000000}+100\left(7560\right)\\ =180\sqrt{9640000}+756000\\ =180\sqrt{{200}^2\cdot 241}+756000\\ =180\times 200\sqrt{241}+756000\\ =36000\sqrt{241}+756000\\ =1314870\end{array}$

Below is a table of distance values for various values of $x$ :

$x\text{ft}$	$C\left(x\right)$
$0$	$1200000$
$500$	$1175812$
$1000$	$\text{1186512}$
$1500$	$1212000$
$2000$	$1243732$
$2500$	$1278479$
$3000$	$1314870$

From $x=0$ to $x=500$ , $C\left(x\right)$ decreases continuously, then continues to increase from $x=1000$ onwards. The minimum cost will be between $x=500$ and $x=1000\text{ ft}$ .

Thus, from point $P$ , the least expensive location appears to be less than $2000\text{ ft}$ away.