Interesting Practical Mathematical Problem

[I_am_learning]

I was making a PCB drilling machine at home with my brother, and came across this mathematical? problem.

I needed to fix the drill machine. But I needed to determine where should it be fixed so that I could drill largest possible PCB boards.
In other words, find the value of x so that you could drill rectangular PCBs of largest possible width.
When I say 'could drill PCB', I mean, you need to be able to place every point of the PCB below the drill so as to be able to drill anywhere in the PCB.

Ask if I didn't make any sense.

The question isn't as simple as it may look.
But its not hard enough to deserve to be posted in maths forum.
You may think you already know the answer when you really don't.
We enjoyed solving this problem, so I shared it here.

 

[dipole]

can you draw a simpler picture that only contains the relevant information?

 

[Jimmy Snyder]

Let A be the distance wall to wall. Place it so that x = A/3. Then you will be able to drill a hole at any point on a board that is 2A/3 or smaller in it's smallest dimension, and have the greatest coverage for larger boards.

 

[genericusrnme]

you could suspend your drill part to the roof so you want work with whatever sized board you want!

pew pew, thinking outside of the box!

 

[I_am_learning]

Jimmy got it right.
Did you get it immediately, or It was only me that needed to spin his head to get the answer.?
We almost fixed it in the middle, before realizing the correct answer.

 

[MarcoD]

We almost fixed it in the middle, before realizing the correct answer.

I still don't see it.

Ah, now I do.

 

[zoobyshoe]

Kudos to Jimmy. My solution was: check Craigslist for a used drill press.

 

[Pythagorean]

x->inf

 

[KingNothing]

Jimmy got it right.
Did you get it immediately, or It was only me that needed to spin his head to get the answer.?
We almost fixed it in the middle, before realizing the correct answer.

Think about it this way: whenever you move the drill in one direction, you are sacrificing room on that side. At 2/3 you are sacrificing the most room while the other side can compensate. If you go further than 2/3 you create a 'dead zone'.

 

[Jimmy Snyder]

Did you get it immediately?
We almost fixed it in the middle, before realizing the correct answer.

No, like you, my first guess was also in the middle. It took me a few moments to realize that you can turn the work around and double the reachable area. My second guess was A/4, but I took a piece of paper and slid it around to see what would happen. That's how I found the answer.

 

[I_am_learning]

second guess was A/4

Mine too. :)

but I took a piece of paper and slid it around to see what would happen. That's how I found the answer.

Quite differently I took a piece of paper and tried to recall linear Programming lessons.