# Thread: Minimization of the distance between 3 points

1. ## Minimization of the distance between 3 points

Centerville is the headquarters of Greedy Cablevision Inc. The cable company is about to expand service to two nearby towns, Springfield and Shelbyville. There needs to be cable connecting Centerville to both towns. The idea is to save on the cost of cable by arranging the cable in a Y-shaped configuation.

Centerville is located at ${\left({11},{0}\right)}$ in the ${x}{y}$-plane, Springfield is at ${\left({0},{5}\right)}$, and Shelbyville is at ${\left({0},-{5}\right)}$. The cable runs from Centerville to some point ${\left({x},{0}\right)}$ on the ${x}$-axis where it splits into two branches going to Springfield and Shelbyville. Find the location ${\left({x},{0}\right)}$ that will minimize the amount of cable between the 3 towns and compute the amount of cable needed.

Not sure how to start on this one. I know that the distance formula is needed, I'm just not sure how.

2. This looks like an optimalization problem. What you need to do is make a function T(x) for the amount of cable that is needed depending on the "branching distance" x from the origin. Then you need to differentiate that formula, and equal it to 0 to find the extremum of the function. (In this case you'll want a minimum, since you're looking for the least possible amount of cable needed).

In this case, the formula for the amount of cable needed would be

$\displaystyle T = 11-x + 2\sqrt{25+x^2}$

In which 11-x is the amount of cable from Centerville to the branching point, and the other term is the amount of cable from the two branches (using Pythagoras).

3. Originally Posted by momopeaches
Centerville is the headquarters of Greedy Cablevision Inc. The cable company is about to expand service to two nearby towns, Springfield and Shelbyville. There needs to be cable connecting Centerville to both towns. The idea is to save on the cost of cable by arranging the cable in a Y-shaped configuation.

Centerville is located at ${\left({11},{0}\right)}$ in the ${x}{y}$-plane, Springfield is at ${\left({0},{5}\right)}$, and Shelbyville is at ${\left({0},-{5}\right)}$. The cable runs from Centerville to some point ${\left({x},{0}\right)}$ on the ${x}$-axis where it splits into two branches going to Springfield and Shelbyville. Find the location ${\left({x},{0}\right)}$ that will minimize the amount of cable between the 3 towns and compute the amount of cable needed.

Not sure how to start on this one. I know that the distance formula is needed, I'm just not sure how.
Take x as being the distance from the y-axis to the connecting point of the cables.
Then the distance from x to Centerville is 11-x.

The distance formula is Pythagoras' theorem in co-ordinate form.

The arms are governed by Pythagoras' theorem

$\displaystyle x^2+5^2=L^2$

$\displaystyle L=\sqrt{5^2+x^2}=\left(5^2+x^2\right)^{0.5}$

On the graph, the tangent will be horizontal at a minimum, hence

$\displaystyle \frac{d}{dx}\left[11-x+2\left(5^2+x^2\right)^{0.5}\right]=0$

$\displaystyle -1+2(0.5)\left(5^2+x^2\right)^{-0.5}(2x)=0$

Be careful that you don't state that the horizontal length plus one arm must be a minimum, as you'd have a straight line from Springfield to Centerville.

$\displaystyle \frac{2x}{\sqrt{5^2+x^2}}=1$

$\displaystyle 2x=\sqrt{5^2+x^2}$

$\displaystyle 4x^2=25+x^2$

$\displaystyle 3x^2=25$

$\displaystyle x=\sqrt{\frac{25}{3}}$

4. Thanks to both of you! I just wasn't sure how to set the problem up. Seems easy now.