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What is the Longest Pole?

Date: 04/04/99 at 21:31:09
From: Marya marinsek
Subject: Pole Length Problem

A metal pole L ft long is pushed on the floor from a corridor 5 ft 
wide into another corridor 4 ft wide and at a right angle. What is the 
longest pipe that can be pushed into the 4-ft corridor?

Date: 04/30/99 at 11:33:05
From: Doctor Fwg
Subject: Re: Pole Length Problem

Dear Marya,

Let's restate the problem: 

   Find the length of the longest rigid (and slender) pole that can be 
   carried around the intersection of two hallways (one of width "X" 
   and the other of width "Y") if the pole axis is kept parallel to 
   the plane of the floor when it turns around the corner:


If one lets the entire pole length = L, then:

   L = L1 + L2,

but:  L1 = Y/[Sin(a)] and L2 = X/[Cos(a)], so:

   L =  Y/[Sin(a)] + X/[Cos(a)].

Now: L is expressed as a function of the hall widths (which are 
constants) and the angle "a" (which is a variable) that occurs when 
the pole is just touching the corner and the intersecting walls of the 
corridors. Substituting varying arbitrary values for the angle "a" 
(ranging between, but not including, 0 and 90 degrees), an extensive 
range of "L" values can be calculated. If one plots these values of L 
as a function of the angle "a", a bowl-shaped curve (or line) will be 
produced (concave upward). The slope of this line is zero at the 
minimum value of L located on this curved line. The minimum value of 
L, at this angle "a", is the length of the longest pole that can just 
turn around the corner that exists due to the intersecting corridors.

The classical technique for finding a minimum (or maximum) along any 
curved line is to take the first derivative of the line function, set 
that result equal to zero, and solve for the minimum (or maximum) 
value of interest. In this case, one can take the first derivative of 
L with respect to the angle "a", set this result equal to zero, and 
solve for the minimum value of L. This minimum in L is, in fact, the 
longest value that L can have which will permit the pole to turn 
completely around the corner. Any pole length longer than this will 
not allow the pole to turn completely and thus pass from one hallway 
to the other. Note also that this procedure should produce a result 
that is independent of travel direction.

Using the equation last written above and following this outline, one 

dL/da = -Y[Cos(a)]/[Sin(a)]^2 + X[Sin(a)]/[Cos(a)]^2 = 0.

Simplifying this equation, one will find that:

Tan(a) = [Y^(1/3)]/[X^(1/3)], therefore:

A right triangle having an abscissa side length of X^(1/3) and an 
ordinate side length of Y^(1/3), as well as angles of "a", "(90-a)", 
and "90" degrees, exists here. Using the Pythagorean Theorem, one can 
show that the hypotenuse (C) of this triangle is:

C = [X^(2/3) + Y^(2/3)]^(1/2), so:

Sin(a) = [Y^(1/3)]/C and Cos(a) = [X^(1/3)]/C.

Now, re-writing L using these expressions:

L = Y/[Sin(a)] + X/[Cos(a)] = Y/[C*Y^(1/3)] +  X/[C*X^(1/3)].

Simplifying and factoring the expression immediately above yields:

L = [X^(2/3) +  Y^(2/3)]^(3/2).

I hope I haven't made any errors - make sure you check this over in 
detail. After checking this solution, see if you can do this problem 
by yourself without looking at the illustration.

With best regards,

- Doctor Fwg, The Math Forum
Associated Topics:
High School Calculus
High School Trigonometry

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