| Re .0:
Consider two points, one the corner and one the point where the ladder
touches the other side of the 8 foot passage. Our initial condition
has one end of the ladder at the corner and the other at its farthest
down the 8 foot passage. The angle from the ladder to the wall is O,
where O = arc sin(8/29).
As we move the ladder up the passage, a bit of it sticks out. Now the
portion of the ladder still in the 8 foot passage is the hypotenuse of
a right triangle, where O is another angle and the length of the
opposite leg is 8 feet. So the length of that portion of the ladder is
8 / sin(O). The length of the rest of the ladder is 29 - 8/sin(O).
The horizontal amount by which that rest of the ladder sticks out
beyond the corner is cos(O) * (29-8/sin(O)).
Take the derivative of that, solve for a zero value between arc
sin(8/29) and a right angle, and put the result back in the above
formula. The length is about 12.685.
-- edp
|
| Suppose we draw this as:
\
\================================
\
\
\
O===============================
The '='s are the walls, and the '\'s are the ladder, and 'O' is the
origin, (0,0).
The ladder has length l, the corridor has width w, one end of the
ladder is at (t,0) (we'll vary t), the ladder is constrained to pass
thru the point (0,w), and the other end of the ladder is at (x,y).
What is the minimum value of x, as t varies from 0 to l?
2 2 2
We have that (x-t) + (y-0) = l , and (x-t)/(y-0) = (0-t)/(w-0),
or (rephrasing), y = w (t-x)/t. Substituting this into the first
equation, we have:
2 2 2 2 2
(x-t) + (t-x) w /t = l
2 2 2 2 2
(x-t) (w + t )/t = l
2 2
x = t � l t / sqrt(w + t )
Now,
2 2 2 3/2
dx/dt = 1 � l w / (w + t )
Setting this to zero, and solving for t, we have in succession:
2 2 2 3/2
1 = l w / (w + t ) (the � must be a minus)
2/3 4/3 2 2
l w = w + t
2 2/3 4/3 2
t = l w - w
2 2/3 4/3 2
t = l w - w
2/3 4/3 2
t = � sqrt( l w - w )
2/3 4/3 2
t = sqrt( l w - w ) (the � must be a plus)
Substituting back into the equation for x, we have:
2 2
x = t - l t / sqrt(w + t )
1/3 2/3 2/3
= t - l t / (l w ) = t [ 1 - (l/w) ]
1/3 2/3 2/3
= t - l t / (l w ) = t [ 1 - (l/w) ]
2/3 4/3 2 2/3
= - sqrt( l w - w ) ( (l/w) - 1 )
2/3 -2/3 2/3
= - w sqrt( l w - 1 ) ( (l/w) - 1 )
2/3 3/2
x = - w ( (l/w) - 1 ) . (a pretty result)
Substituting l = 27, and w = 8, (interesting numbers!) we get:
2/3 3/2 3/2
x = - 8 ( (27/8) - 1 ) = - 8 ( 9/4 - 1 )
3/2 3/2
= - 8 (5/4) = - 8 ( 5 / 8 )
3/2
x = - 5 = - 11.1803...
|
| Here's a related problem that is somewhat easier to solve:
A building has two hallways meeting at right angles, each 6 feet wide. If all
sofas are 3 feet wide, what's the longest sofa that can be moved from one
hallway to the other (while being kept level)?
In this version you can take advantage of symettry to simplify the
solution. Only quadratics are involved.
Lynn Yarbrough
|
| No, no, I said *level*. No fair tipping it.
Actually, the length of sofa in this problem actually exceeds the height of
most ceilings (8 ft.) but let's leave the 3-d case as an exercise for the
ambitious reader.
Lynn
|