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Carrying a Ladder around a Corner

Date: 02/28/2003 at 23:31:12
From: Scott
Subject: Carrying a Ladder around a Corner

A ladder of length L is carried horizontally around a corner from a 
hall 3 feet wide into a hall 4 feet wide. What is the length of the 
ladder?

  3 ft.
+\\   +
+ \\  +
+  \\ +
+   \\+ + + + + +
+    \\ )theta
+     \\         4 ft.
+      \\
+ + + + + + + + +


No angle is given.

I assumed theta to be 45 degrees and split the problem into two. After 
using arctan and the Pythagorean theorem I got L=10 ft. The reason I 
used 45 degrees is that the ladder was sitting between a halls that 
are at 90 degrees.


Date: 03/10/2003 at 11:40:55
From: Doctor Marshall
Subject: Re: Carrying a Ladder around a Corner

Hi Scott, thanks for writing to Dr. Math.

What is this question for, a real life scenario or a math problem you 
found somewhere? I ask because the solution is deceptively 
complicated.

Also, do we need to consider the thickness of the ladder?  I'll assume
at first that we don't. 

Suppose we can find a function, L(theta), which returns, for any given
value of theta, the length of the longest ladder that can fit. Then
what we want to find is the value of theta for which L(theta) is a
minimum. A ladder with that length will fit at any angle; and for any
ladder longer than this, there will be some value of theta for which
the ladder won't fit.  

The picture:

    3 ft.
  +\\     +                 
  + \\    +
  +  \\   +
  +   \\  +
  +    \\ +
  + _t(_\\+ + + + + +
  +      \\ )t                             t = theta
  +       \\         4 ft.
  +        \\
  +         \\
  +          \\
  + + + + + + + + + + 

can be disassembled to form two triangles:

  |\                  ------+
  | \                 \t    |
  |  \h                \    |
  |   \                 \   |4
  |  t(\                i\  |
  +-----                  \ |
    3                      \|

If h and i are the hypotenuses, then 

  i = 4/sin(t) 

    = 4csc(t)

and

  h = 3/cos(t) 

    = 3sec(t)            

So we can express the relevant function as

  L(t) = i + h

       = 3sec(t) + 4csc(t)             

When given a value for theta, the function returns the length of the
ladder that will just fit between the walls. 

Now what we want to do is find the derivative of L with respect to t,
and use that to find the value of t for which L is a maximum. Any
maximum must occur either at an endpoint of the domain (i.e., the set
of legal values for t), or when the derivative is zero.  

Alternatively, we could simply plot L(t), and use the graph to
determine the maximum value of L, and the value of t that corresponds
to it. 

Note again that we've neglected to take the thickness of the ladder
into account. Doing that would make the problem messier.

Write back if you have any other questions, or need something 
explained.

- Doctor Marshall, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Geometry
High School Practical Geometry
High School Triangles and Other Polygons
High School Trigonometry

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