Randomly Cutting a Rope into Two Segments

Date: 05/14/98 at 20:12:26
From: William Luh
Subject: Probability of cutting ropes 

A rope 20m long is randomly cut into two segments, each of which is 
used to form the perimeter of a square.

   (a) Find the probability that the larger square has an area greater 
       than 9m^2.

   (b) Find the probability that the total area of the two squares is 
       greater than 20.5m^2.

The reason why I am stumped is because the rope can be cut to any 
length. Essentially, I can have 0.01m and 19.99m.

Here's what I've tried:

a) Let piece 1 = p
   Let piece 2 = 20 - p       restrictions 0 < p < 20

Then make squares with pieces 1 and 2:

   Area of square for piece 1: (p/4)^2
   Area of square for piece 2: ((20-p)/4)^2

Let piece 1 be the larger of the two, so:

    p > 20 - p
   2p > 20
    p > 10

This makes sense. If p = 10, then both squares would have the same 
dimensions, and we would not be able to determine a LARGER square from 
the two because they would be equal.

Now, for part a), we are told that the larger area has to be greater 
than 9m^2. So:

   (p/4)^2 > 9
       p^2 > 144
        p  > 12

To have an area greater than 9 units square, the perimeter or length 
of the piece would have to be greater than 12 units.

So the question asks for the probability, and I answer:

   P(Area > 9|larger square)
   = P(Area > 9 AND larger square)/P(larger square)
   = 7/9

I'm trying to decide how to interpret the inequalities. Should the 
probability be 7/9 or 8/10? I'm confused.
==============================================================

Part (b) is also interesting, because I used some calculus.

I will restate the question now:

   Find the probability that the total area of the two squares is 
   greater than 20.5m^2.

Let function A(p) represent the sum of the areas, so:

   A(p) = (p/4)^2 + ((20-p)/4)^2    using info from first part

The graph of the function A(p) is a parabola concave up, so we will 
find a minimum area if we set A'(p) = 0:

   A'(p) = 2(1/4)(p/4) + 2(-1/4)((20-p)/4)

Set A'(p) = 0 and solve for p. The result is p = 10, as predicted. 
When p = 10, A(10) = 12.5.

The maximum would occur when one piece of the string approaches 0 
while the other approaches 20m, yielding a maximum area of 25m^2.

So:        

   12.5 <= A(p) <= 25

The question asks to find probability that A(p) > 20.5.

If we work through the mechanics of A(p) > 20.5, we find that this 
occurs when:

   0 <  p < 2       (1)
   18 < p < 20      (2)

Now we must remember that the maximum area occurs when p -> 20 and 
minimum area occurs when p = 10, so there is a 10 unit range.

Looking at restriction (2), there is a 2 unit range. Therefore, the 
probability is 2/10 = 0.2

Please help me, Dr. Math. I know there is something missing or 
something flawed, but I can't seem get my brain around this problem. 
If I'm totally on the wrong track, please show me how to do at least 
part (a) or (b).

Thanks,
William

Date: 05/15/98 at 10:58:56
From: Doctor Pat
Subject: Re: Probability of cutting ropes 

William,

Wow, you put a lot of good work into this, and I think you got the 
second part exactly right. In the first part you did some great 
analysis and walked right up to the answer, but didn't see it.  

I would suggest thinking of the problem as making a single cut, 
instead of making two pieces. One thing is easier to focus on than 
two, usually. Think of the rope as laying along the x-axis from zero 
to 20. If you use your value p as the coordinate of the cut, then one 
of the pieces is p units long, and the other is (20 - p), as you also 
figured. Now revisit the problem thinking of this question:  

   If I picked a point at random along the segment from 0 to 20 to 
   make the cut, where would it have to be to produce a square with an 
   area greater than 9 sq. units?

You have really found this answer already. Just figure what fraction 
of the time this would occur by chance, and I think you will see the 
solution. 

Good luck!

-Doctor Pat, The Math Forum
Check out our web site! http://mathforum.org/dr.math/   

Date: 05/16/98 at 18:47:51
From: William
Subject: Re: Probability of cutting ropes 

Dear Dr. Pat,

Thank you for the encouragement, and finding my mistake. I have 
revised part (a). I think the probability is conditional because we 
are asked to examine the large square and not just any square. I took 
your advice and drew a number line:

   ------------------------------------------------------------------
   0                                 10        12                  20

   |----------------------------------|  P(larger square)

   |---------------------------|         P(area > 9 AND larger square)

Therefore, IF this is indeed a conditional probability, then:

P(area>9|larger square) = P(area>9 AND larger square)/P(larger square)
                        = (8/20) / (10/20)
                        = 4/5

Would this be correct now?

Thank you,
William

Date: 05/19/98 at 16:02:31
From: Doctor Anthony
Subject: Re: Probability of cutting ropes 

For an area of 9m^2, the side must be greater than 3, so perimeter 
will be greater than 12. If you have a 20m length of rope, then a cut 
in the range 0 < x < 8 or in the range 12 < x < 20 will give you one 
piece with a perimeter greater than or equal to 12.

The probability of a cut in one or the other of those regions 
(assuming the cut to be equally likely at any point) is 16/20 =  4/5.

If the cut is at a distance x from one end, the two portions are of 
length x and 20 - x. The sides are of length x/4 and (5 - x/4), and 
the areas total to:

    A = x^2/16 + (5 - x/4)^2 > 20.5

        x^2/16 + 25 - 5x/2 + x^2/16 > 20.5

        x^2/8 - 5x/2 + 9/2 > 0

        x^2 - 20x + 36 > 0

        (x - 2)(x - 18) > 0

This will be satisfied if x < 2  or  x > 18

So probability of being in either of those two ranges is 4/20 = 1/5.

So probability that area is greater than 20.5 is 1/5.
                  
-Doctor Anthony, The Math Forum
Check out our web site! http://mathforum.org/dr.math/