1. Aug 2, 2012

### unscientific

1. The problem statement, all variables and given/known data

The problem is about finding the longest ladder length L that can be carried horizontally across a corner between 2 corridors.

3. The attempt at a solution

I understand that the solution's using the Lagrange method but what I don't get is why they can assume that at (a+δ, 0) and (0, b+η) are the points of the longest ladder that touch the sides.

I attached a picture to demonstrate an extreme case of what I'm talking about.. (It satisfies the relation ab = ηδ by comparing tan θ.

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2. Aug 2, 2012

### jackmell

Suppose I already had the longest ladder and I start bringing it around. It will just barely fit when it touches the corner of the inside wall right? Then surely at that point, the ends will touch the outside walls along the x and y axis at a+delta and b+eta right? Actually the points a+delta and b+eta are the points along the outside wall where any size ladder touches. Maybe though I'm not understanding what you're asking. Sorry and I can't delete this.

Last edited: Aug 2, 2012
3. Aug 2, 2012

### unscientific

Hmm what I'm asking is: how can you assume that the longest ladder will "barely" touch the corner and the other 2 points on the walls?

In the picture attached I have already shown that an extremely long ladder that satisfies the conditions but obviously cant be brought through.

My point is that the conditions aren't stringent enough.

4. Aug 2, 2012

### jambaugh

This is a necessary but not sufficient condition. Clearly a ladder cannot be longer than one which "just barely touches". To solve the problem find the position at which such a "just touching ladder" is shortest in length. (One can assume by symmetry this occurs at the 45deg mark but work it out via optimization methods since that is the point of the exercise.)

5. Aug 2, 2012

### clamtrox

You are correct. There is a small typo - you are trying to find the minimum ladder length satisfying that constraint. This is easy to see by calculating $\frac{\partial^2 f}{\partial \xi^2} = \frac{\partial^2 f}{\partial \eta^2} = 2 > 0$, meaning the only extremum is a minimum, not a maximum.

6. Aug 2, 2012

### HallsofIvy

Staff Emeritus
if it doesn't "barely touch" then either:
1) it doesn't fit!
or
2) A slightly longer ladder will fit around the corner.

7. Aug 2, 2012

### jackmell

Isn't that intutitive? Consider a small one first. Surely it's going around the corner without touching all three points (the inner corner and the outside walls). Now consider one that's too long. It's touching all three points but not going around. So the longest one I can get around is slightly shorter than one that touches all three points.

Or actually the longest is one that goes around and just touches all three points.

8. Aug 2, 2012

### voko

In fact, you could make no assumptions on what the ends of the ladder touch. But then you would need to formulate conditions that allow no part of the ladder within the walls. Those are more complex conditions - involving inequalities - but ultimately you would get the same result.

9. Aug 2, 2012

### unscientific

1. Does it matter at what angle the wall just nice touches the corner (a,b) ?
2. If so, from what angle onwards will the ladder not be able to pass through?
3. How can you tell when a ladder is "just long enough" for it to pass through?

10. Aug 2, 2012

### jackmell

I made a mistake. It's not so intutitive until after I think about it a while and study the example in my book. Sorry about that. Also, don't forget what Clamtrox said. There is a mistake in the handout. You are trying to "minimize" the length of a line drawn when you touch the outside walls and touch the inner corner. Think first about lines which are too long and don't get through. Draw some of those. Then draw smaller ones, eventually you reach a largest line that does get through (while still touching at the three indicated points). That minimum value gives the maximum length of the ladder that will just get through. You can also work this in terms of the angle the line makes with the walls then express the length of the line as a function of the angle it makes, then minimize the length as a function of the angle.

Last edited: Aug 2, 2012
11. Aug 3, 2012

### unscientific

Yes then differentiate set to zero..that method works. The method presented in this book is the lagrange multiplier. I've given some thought about it and I think the lagrange multiplier is in fact, quite an elegant solution as it does away with some of the questions above:

Let's first set the assumptions of transporting the ladder (without this definition, you can't work anything out):

1. The ladder must, at all times touch both ends of the wall as it turns (angle changes decreases from 90)

2. It is trivial to see that at all points during its motion, if the ladder never encounters the corner, the ladder is brought across successfully.

3. Hence, we want to find the shortest possible length, across all angles during transportation (remember assumption 1. holds) that the ladder will just nice touch the corner.

Lagrange multiplier
1. As you examine the conditions, it merely stated that the ladder must touch all three points (both walls and corner)

2. By finding the minimum length to do so, it eliminates the question of finding at what angle the ladder is most likely to hit the corner or when during the turning point.

3. The constraints work for all angles (that's the beauty, isn't it?) when you simply define the y=mx+c equation for the straight line.

So in short, with the lagrange method, you are simply finding across all angles, for any a and any b the shortest ladder length to satisfy the conditions above.