Expert: Paul Klarreich - 6/24/2008

Question
Find the longest beam that can go around a hallway with width A to a hallway with a width B.

Tried everything, Obviously involves similar triangles and some trig functions, but I cannot find a significant relationship. Looking to find the MAX length that La + Lb (hypotenuse of the largest triangle) can be through optimization I believe, don't quote me on that. I think the picture is an accurate representation.

Answer
Questioner:   Willis
Category:  Calculus
Private:  No
 
Subject:  Optimization
Question:  Find the longest beam that can go around a hallway with width A to a hallway with a width B.

Tried everything, Obviously involves similar triangles and some trig functions, but I cannot find a significant relationship. Looking to find the MAX length that La + Lb (hypotenuse of the largest triangle) can be through optimization I believe, don't quote me on that. I think the picture is an accurate representation
.............................................
Hi, Willis,

>> What picture?

Ah, well, it probably looks like this:

(I remember this from some years ago.  Alas, I cannot find the book that had it -- it's probably the old Thomas book from the 1950's.)

There probably is an easier way, but this is the best I can come up with.


  |<a->|
  |    |
  |    |
y0 |\   |
  | \  |P(a,b)
  |   \+---------------------
  |    \                    ^
  |     \                   b
  |       \                 v
  +--------------------------    
           x0

PQ = sqrt(a^2 + b^2) = d.

Let t (theta) be the angle with the lower wall.

r^2 = x^2 + y^2, to be maximized.


You have a line that passes through the point (a,b) and has an x-intercept, which we will call, er,..  x0.
              (x0 - a)
Its slope is  ---------
              (0 - b)

  (x0 - a)
=  ---------
    - b

  a - x0
=  ------
    b
                         a - x0
Its equation is  y - b = --------(x - a)
                           b

         a - x0
y - b = --------(x - a) + b
            b

    (a - x0)(x - a) + b^2
y = ---------------------
             b

Now the y-intercept is:

    (a - x0)(0 - a) + b^2
y0 = ---------------------
             b

    (a - x0)(- a) + b^2
y0 = ---------------------
             b

    -a^2 + a x0 + b^2
y0 = ---------------------
             b

Now the length of the segment is  x0^2 + y0^2  [the length^2, actually]
and we have to MINIMIZE this.  (Yes, minimize, because if the beam is longer than the minimum, it won't go around.)

x0^2 + y0^2  =
       [-a^2 + a x0 + b^2]^2
x0^2 + ---------------------
                b^2

Differentiate w.r.t. x0:

       2[-a^2 + a x0 + b^2](a)
2 x0 + -----------------------
                b^2

       2a[-a^2 + a x0 + b^2]
2 x0 + ----------------------
                b^2

Set that = 0:

       2a[-a^2 + a x0 + b^2]
2 x0 + ---------------------- = 0
                b^2


       a[-a^2 + a x0 + b^2]
x0 + ---------------------- = 0
                b^2

    
b^2 x0 - a^3 + a^2 x0 + ab^2 = 0

x0(a^2 + b^2) = a^3 - ab^2

x0(a^2 + b^2) = a(a^2 - b^2)
     a(a^2 - b^2)
x0 = -------------
     (a^2 + b^2)

I think you can handle the rest.