Chapter 1.2, Problem 56E

(a)

To determine

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

Answer to Problem 56E

The cost of laying the cable in terms of distance is given by the function $C\left(x\right)=180\sqrt{{800}^2+x^2}+100\left(10560-x\right)$ .

Explanation of Solution

Given information:The width of a river is 800 ft. The cost for laying a cable from the plant to a location to the opposite side in the city 2 mi downstream is $180 per foot across river and $100 per foot along the land.

The cable goes through the power plant to Q on opposite side x ft from point P .

  

Calculation:

Let the length of the cable from power plant to point Qbe $y_1$ . Use Pythagoras theorem to find the length of cable $y_1$ .

  $\begin{array}{c}{y}_1{}^2=x^2+{800}^2\\ y_1=\sqrt{x^2+{800}^2}\end{array}$

It is known that $2\text{mi}=10,560\text{ft}$ .

Let the length of the cable that along the land be $y_2$ . So,

  $y_2=10560-x$

The function for laying the cable across the river and along the land is:

  $\begin{array}{l}C\left(x\right)=180y_1+100y_2\\ C\left(x\right)=180\sqrt{{800}^2+x^2}+100\left(10560-x\right)\end{array}$

Therefore, the cost of laying the cable in terms of distance is given by the function $C\left(x\right)=180\sqrt{{800}^2+x^2}+100\left(10560-x\right)$ .

(b)

To determine

To find:The least expensive location for point Q is less than 2000 ft or greater than 2000 ft from the point P .

Answer to Problem 56E

The least expensive location for point Q is less than 2000 ftfrom the point P .

Explanation of Solution

Given information:The width of a river is 800 ft. The cost for laying a cable from the plant to a location to the opposite side in the city 2 mi downstream is $180 per foot across river and $100 per foot along the land.

  

Calculation:

From part (a), the function that gives the cost of laying the cable in terms of distance is $C\left(x\right)=180\sqrt{{800}^2+x^2}+100\left(10560-x\right)$ .

Substitute the different values of x and make a table for the cost.

x	$C\left(x\right)$
0	$1,200,000
500	$1,175,812
1000	$1,186,512
1500	$1,212,000
2000	$1,243,732
2500	$1,278,479
3000	$1,314,870

From the table it can be observed that the cost for lying a cable is least expensive when the location for point Q is less than 2000 ft from the point P .

Therefore, the least expensive location for point Q is less than 2000 ft from the point P .