# Thread: minimum amount of pipe

1. ## minimum amount of pipe

We've seen the problems where the minimum amount of pipe is asked for, but here's one a wee bit trickier if anyone would like a go. Nice problem.

"Two houses are located along a straight road, 300 yards apart. A well is to be dug 400 yards from and perpendicular to the main road at a point between the houses, 200 yards from one of the houses. They plan to buy some 1 inch pipe to lay from the well to a junction box and use 400 yards of 3/4 inch pipe they already have to go from the junction to each house.
Where should the junction be located to minimize the amount of 1 inch pipe they need to buy?."

Perhaps someone has a cool approach. I will post my method later.

2. I tried doing this by coordinate geometry, taking the origin as the midpoint between the houses. Then the houses are at the points (±150,0) and the well is at (50,400). If the junction is at (x,y) then the sum of the distances from (x,y) to the houses is 400. So (x,y) lies on an ellipse with foci at the houses, major axis 400 and eccentricity 3/4. The equation of the ellipse is $\displaystyle \frac{x^2}{16} + \frac{y^2}{7} = 2500$, and we want to find the shortest distance from (50,400) to a point on the ellipse.

That's where I get stuck. The shortest distance seems to be given by a quartic equation with horrendous coefficients, and I can see no way of solving it. A bit of numerical trial and error seems to show that the minimum distance is about 269.9 yards, when the junction is somewhere near the point (26,131).

Is there some trick here that I'm missing, or have I made some silly mistake?

3. By Jove, Opalq, that's how I done it...with the ellipse. I was wondering if I was correct or not. We got the same answers, except I got 23.6 for the x coordinate.

Here's what I done(same as you):

The junction point must be on the ellipse. with foci (-100,0), (200,0).

And vertices (-150,0) and (250,0).

This gives the ellipse equation of

$\displaystyle \frac{(x-50)^{2}}{(200)^{2}}+\frac{y^{2}}{7(50)^{2}}=1$

The length of pipe between the well and the junction is $\displaystyle D=\sqrt{x^{2}+(y-400)^{2}}$

If we solve the ellipse equation for y and sub into D, we get an awful looking thing:

$\displaystyle D(x)=\sqrt{x^{2}+\left(50\sqrt{7}\sqrt{1-\frac{(x-50)^{2}}{(200)^{2}}}-400\right)^{2}}$

Now, this leads us to a minimum of 269.9 when (x,y)=(23,6, 131.1)

I will admit, I used my calculator to do the rough stuff.