Chapter 7.6, Problem 37E

To determine

To find: the lowest possible cost of connecting the existing water mains to the new tower.

Answer to Problem 37E

$6927.82

Explanation of Solution

Given information:

The equations are as follows:

  $\begin{array}{c}5x-y=15\\ 3x+4y=36\\ 5x-2y=-20\end{array}$

Water tower is placed at origin.Cost of laying new underground pipe per 100 feet is $500.

Calculation:

Convert each line equation in normal form, i.e. $x\cos \varphi +y\sin \varphi -p=0$ .

  $\begin{array}{c}5x-y=15\left[\text{Subtract 15 from each side}\right]\\ 5x-y-15=0\left[\text{Here }A=5,B=-1\text{ and }C=-15\right]\\ \frac{5}{\sqrt{5^2+{\left(-1\right)}^2}}x-\frac{1}{\sqrt{5^2+{\left(-1\right)}^2}}y-\frac{15}{\sqrt{5^2+{\left(-1\right)}^2}}=0\left[\text{Divide each side by }\sqrt{5^2+{\left(-1\right)}^2}\right]\\ \frac{5}{\sqrt{26}}x-\frac{1}{\sqrt{26}}y-\frac{15}{\sqrt{26}}=0\\ \text{Here }{p}_1=\frac{15}{\sqrt{26}}\\ 3x+4y=36\left[\text{Subtract 36 from each side}\right]\\ 3x+4y-36=0\left[\text{Here }A=3,B=4\text{ and }C=-36\right]\\ \frac{3}{\sqrt{3^2+4^2}}x+\frac{4}{\sqrt{3^2+4^2}}y-\frac{36}{\sqrt{3^2+4^2}}=0\left[\text{Divide each side by }\sqrt{3^2+4^2}\right]\\ \frac{3}{5}x+\frac{4}{5}y-\frac{36}{5}=0\\ \text{Here }{p}_2=-\frac{36}{5}\end{array}$

  $\begin{array}{c}5x-2y=-20\left[\text{Add 20 on each side}\right]\\ 5x-2y+20=0\left[\text{Here }A=5,B=-2\text{ and }C=20\right]\\ \frac{5}{-\sqrt{5^2+{\left(-2\right)}^2}}x-\frac{2}{-\sqrt{5^2+{\left(-2\right)}^2}}y+\frac{20}{-\sqrt{5^2+{\left(-2\right)}^2}}=0\left[\text{Divide each side by }-\sqrt{5^2+{\left(-2\right)}^2}\right]\\ -\frac{5}{\sqrt{29}}x+\frac{2}{\sqrt{29}}y-\frac{20}{\sqrt{29}}=0\\ \text{Here }{p}_3=\frac{20}{\sqrt{29}}\end{array}$

Since the value of $p$ are in hundred feet, then multiply $500 to the total distance to evaluate the lowest possible cost can be evaluated as follows:

  $\begin{array}{c}C=\$500\left(p_1+p_2+p_3\right)\\ =\$500\left(\frac{15}{\sqrt{26}}+\frac{36}{5}+\frac{20}{\sqrt{29}}\right)\left[\text{Put the value of }{p}_1\text{, }{p}_2\text{ and }{p}_3\right]\\ =\$500\left(\frac{15\sqrt{26}}{\sqrt{26}\left(\sqrt{26}\right)}+\frac{36}{5}+\frac{20\sqrt{29}}{\sqrt{29}\left(\sqrt{29}\right)}\right)\\ =\$500\left(\frac{15\sqrt{26}}{26}+\frac{36}{5}+\frac{20\sqrt{29}}{29}\right)\\ =\$6927.82\end{array}$

Thus, the lowest possible cost of connecting the existing water mains to the new tower is $6927.82.