What have you done so far?

Clarification question: is that $12/m or $12/mm for the underwater cost?

Results 1 to 9 of 9

- June 21st 2010, 08:08 AM #1

- Joined
- Jun 2010
- Posts
- 4

## Optimation Problem

A house (H) is located on an island that is 200 north of a point (A) on the shore. A Power station (p) is located on the shoreline at a point 400 m west of Point A. An electric cable is going to be laid from the power station to the house. It costs $12/m to lay underwater and $6/m to lay underground. How should the cable be placed so that the cost is a minimum and what is this minimum cost?

Please help, exam is in 1.5 hour. Thanks

- June 21st 2010, 08:11 AM #2

- June 21st 2010, 08:12 AM #3

- Joined
- Jun 2010
- Posts
- 4

- June 21st 2010, 08:15 AM #4
Well, to start you out, I'd go with

,

where is the length of underwater cable, and the length of underground cable.

So, I'm assuming you're going to want to go some distance east from the power station, underground, and then at some point you're going to go underwater. Is this correct?

- June 21st 2010, 08:18 AM #5

- Joined
- Jun 2010
- Posts
- 4

- June 21st 2010, 08:20 AM #6

- June 21st 2010, 08:36 AM #7

- Joined
- May 2010
- Posts
- 1,034
- Thanks
- 28

- June 21st 2010, 08:38 AM #8

- Joined
- Jun 2010
- Posts
- 4

- June 21st 2010, 11:38 AM #9

- Joined
- Apr 2005
- Posts
- 17,588
- Thanks
- 2235

I like your disclaimer! And this is not accurate- there are an infinite number of possible routes and neither of the ones you give is likely to be optimal. Put the cable from H to some point X between H and A. Let "x" be the distance from H to X. The cost of laying that cable is 6x.

The distance remaining from X to A is 200- x and the distance across from A to P is 400 so you can use the Pythagorean theorem to find the straight line distance form X to P. Multiply that by 12 to find the cost of laying that cable.

The total cost is the sum of those two costs. Differentiate that sum with respect to x and set the derivative equal to 0 to find the x that gives optimal cost.

By the way, the correct word is "optim**iz**ation", not "optimation".