Optimization question help...?

Question

I really need help the question is that an island is 85m from a straight shoreline at the closest point and the nearest electrical connection is 250m along the shore from that point. It costs twice as much to install cable across water as across land what would be the least expensive way to install cable? Any ideas on how to start off the question would be appreciated.

Answer 1

First, you need to be able to visualize what is happening. Call the point for the electrical connection point E.
Call the point on the island point S.
Call the point on the shore 85 m from the island point R.

Now the wire will NOT run all the way from E to R, but rather, at some point P in between it will turn and start to run diagonally out to R.


Now let x = the distance from P to R.
Then 250 - x = the distance from E to P.

Notice you have a right triangle formed (triangle SRP) so you can use the pythagorean theorem to find the hypotenuse (length SP)

Also, since it costs twice as much to lay the wire under the water as along the shore, we effectively need to double this distance before adding it to the distance from P to E in the optimization problem.

So find the minimum of:

2*sqrt(85^2 + x^2) + (250 - x)


Hope this helps!

Answer 2

If you run the cable straight across, it's 85m in the water, then 250m along the shoreline.

If you run the cable at an angle t from straight across, you'll have the cable running as the hypotenuse of a right triangle where the adjacent side is 85m. Then it's 85m / cos(t) in the water. That point will touch the land at 85m * tan(t) meters closer to the outlet along the shoreline. So it is 250m - ( 85m * tan(t) ) meters across the land from that point.

Then, your cost as a function of angle t, is (ignoring the cost-per-unit, just using twice as much for the water cost as the land cost):

c(t) = 2 * (water) + land
c(t) = 2 * ( 85 / cos(t) ) + 250 - 85 * tan(t)
c(t) = 170*sec(t) + 250 - 85*tan(t)

Now you take the derivative dc/dt, and find where it is equal to zero.

dc/dt = 0
d(170*sec(t) + 250 - 85*tan(t))/dt = 0
170*sec(t)*tan(t) - 85*sec(t)*sec(t) = 0

Divide by 85 sec(t) since sec(t) is never equal to zero and 85 is a common factor:

2*tan(t) - sec(t) = 0
2*tan(t) = sec(t)
2*sin(t)/cos(t) = 1/cos(t)
2*sin(t) = 1
sin(t) = 1/2
t = sin^-1(1/2)
t = 30 degrees

Let's assume $1 per meter on land and $2 per meter on water.

The line straight across is 85*2+250 = $420.
A line at 45 degrees would be 85*2*sqrt(2) + 250 - 85 = $405.42

A line at 30 degrees would be 85*2/cos(30) + 250 - 85*sin(30) = $403.80.

So, we do know that 30 degrees is cheaper than 45 degrees or straight across (0 degrees).

Answer 3

Through land and shortest distance across water will cost
(85*2 + 250 ) = 420

Straight line across water will cost
((85^2 + 250^2) ^ (1/2) )*2= 528

so you decide.

Answer 4

some SMO recent options are as:- • Google Plus One • weblog creation, Posting and advertising • Social Networking Profiles Creations • facebook advertising and marketing • Twitter advertising and marketing • LinkedIn advertising and marketing • Video Submission • Squidoo Lens creation • Hub Pages creation • Use of search engine optimization equipment

Answer 5

I have already answered this one.