Caculating minimum cost of running cable to a house.

Question

If it cost 2.1 cents per foot to run cabel along a road and 3.8 cents per foot to run cable off the road. what is the minimun cost to run cable if a house is two miles from the road and the cable connection is 5 miles from the turn off to go to the house.

Answer 1

I think the previous answer is incomplete, as it determines only the cost of running along the road for the 5 miles, then off for 2.  However, that may not be the minimum cost, because it is possible to run partly along the road and partly off it at a slant.

 

The solution I have for this really depends on a drawing, and, if you would like to see it, I could try to email it to you.  Basically, I defined two variables

 

a = distance cable runs along the road (in miles)

b = distance cable runs off road (in miles)

 

The total cost (in U.S. dollars) of running the cable is then:  cost = a(0.021)+b(0.038)

 

We need to get this expression in terms of one variable on the right.  This is where the drawing would come in handy, because it turns out (by triangle geometry and the Pythagorean theorem) the relationship is:

 

b = √((5-a)²+4) = √(a²-10a+29)

 

So cost = a(0.021)+(√(a²-10a+29))(0.038)

 

To minimize the cost, I take the derivative of cost with respect to a (the distance run along the road), and set that derivative equal to zero to find an extremum for cost.  I expect that extremum to be a minimum, but I can check that later.  

 

Taking the derivative of cost with respect to a involves the chain rule and some algebra (which I can talk to you about in more detail, if you like), but it comes down to:

 

d(cost)/da = 0.021 + (0.038)[(a-5)/√(a²-10a+29)]

 

Now I need to obtain the solution for a that makes d(cost)/da = 0.  This involves setting d(cost)/da = 0 and then using a LOT more algebra to put the previous equation into the standard form of a quadratic equation in a.  Eventually, I got it to be:

 

2.27a²-22.7a+52.75 = 0

 

I was able to solve this for a using the quadratic formula, and ended up with two solutions:

 

a = 3.67 and a = 6.33

 

I can discard the second solution as extraneous, since it exceeds 5 miles, and I know that will cost more (it would be over $700 just to go along the road this distance).  That leaves a = 3.67 miles.

 

Then I go all the way back up to my first equation, which gave b in terms of a, and solve for b.  I get b = 2.31 miles.

 

The total cost for going 3.67 miles along the road and then 2.31 miles off road is (1 mile = 5280 feet):

 

cost = (3.67*5280*0.021)+(2.31*5280*.038) = $870.40  (note that this is definitely below the previous answer)

 

To be complete, I found the second derivative of cost with respect to a and confirmed, for a = 3.67, it was positive, meaning this extremum is a minimum.

 

If you want more details, I am available for online tutoring.  I cannot guarantee all the algebra here (there was a lot of it), and thus I cannot guarantee the final answer.  But I think the technique is sound.

 

Answer 2

 

Answer removed due to a lower cost computed by another tutor.