Question

You have been asked to determine where a water works should be built along a river between Chesterville and Denton to minimize the total cost of the project. The pipe to Chesterville costs $3000 per mile and the pipe to Denton costs $7000 per mile. Find the length of each pipe so that the total cost is a minimum. What is the cost?

Answer 1

Length of pipe to Chesterville is 8.376 miles and

Length of pipe to Denton is 5.46 miles

Explanation:

Here we have

The distance of Chesterville from the river is 3 miles, while the distance of Denton from the river is 5 miles

The bank of the river is 10 miles long

Therefore, we have

If x is the distance from the point directly opposite to Chesterville to the location of the water works, the equation is;

Cost to Chesterville =
`3000* √(x^2 + 3^2)`

Cost to Denton =
`7000* √((10-x)^2 + 5^2)`

Total cost is then;

`7000* √((10-x)^2 + 5^2) + 3000* √(x^2 + 3^2)`

We differentiate the above equation and equate it to zero to get the minimum cost as

`\frac{\mathrm{d} (7000* √((10-x)^2 + 5^2) + 3000* √(x^2 + 3^2))}{\mathrm{d} x}` = 0

`7000(2x-20)/(2√(x^2-20x+125) ) +3000(2x)/(2√(x^2+9) ) = 0`

`3500(2x-20)/(√(x^2-20x+125) ) = -1500(2x)/(√(x^2+9) )`

`3500(√(x^2+9))/(√(x^2-20x+125) ) = -1500(2x)/(2x-20 )`

`x^4-20x^3+10.54x^2-22.05x+110.25 =0`

Solving the quartic equation we get

x = 7.82 miles

Therefore the length of is given as

Length of pipe to Chesterville
`\sqrt{7.82^(2) +3^2 } = 8.376 \, miles`

Length of pipe to Denton =
`√((10-7.82)^2 + 5^2) = 5.46 \, miles`.