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Longest Ladder

Date: 11/30/2001 at 11:44:30
From: Michelle Rittenhouse
Subject: Ladder Problem

Two hallways meet to form a right angle. One is 8 ft. wide and the 
other is 4 ft. wide. What is the longest ladder that can go around 
the corner where the hallways meet? The ladder is carried 
horizontally. Disregard the width of the ladder.

Date: 11/30/2001 at 12:37:41
From: Doctor Rob
Subject: Re: Ladder Problem

Thanks for writing to Ask Dr. Math, Michelle.

I drew this diagram:

      A       x+4      C
      :\              |
      : \             |
      :  \            |
     8:   \           |
      :    \          |
      :     \         |
      :   x  \        |
   ---o-------o       |8+32/x
      E      D|\      |
              | \     |
              |  \    |
          32/x|   \   |
              |    \  |
              |     \ |
              |      \|
            F o.......o B
              |   4   |

Let the length of DE be x. Using similar triangles ADE and DBF, you 
will find that DF = 8*4/x = 32/x. Then using the Pythagorean theorem 
on triangle ABC, the length L of AB is given in terms of x by

   L = sqrt([x+4]^2+[8+32/x]^2),
     = sqrt([x+4]^2*[1+(8/x)^2]),
     = sqrt(x^2+64)*(x+4)/x.

To find the minimum value of L as the length x changes, which is the 
maximum length the ladder can have, you can differentiate this with 
respect to x, set that equal to zero, and solve for x. The equation I 
got after simplifying was

   (x^3-256)/(x^2*sqrt[x^2+64]) = 0.

Solve this for x, and then you can compute the corresponding value of 
L. My answer came out between 15 and 20 feet.

Feel free to write again if I can help further.

- Doctor Rob, The Math Forum
Associated Topics:
High School Calculus
High School Geometry
High School Practical Geometry
High School Triangles and Other Polygons

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