Sent (to Dad by E-mail) September 9, 2000 at 12:11 PM.

Problem Statement:
1.) Over the top of a fence there is placed a rope, the same amount on both sides. The rope
weighs 1/3 lb. per foot. On one end of the rope hangs a monkey holding a banana, and on the
other end is a weight equal in weight to the weight of the monkey.
2.) The banana weighs two ounces/inch. The length of the rope (in feet) is as long as the age of
the monkey (in years), and the weight of the monkey in ounces is as much as the age of the
monkey's mother. The combined ages of the monkey and its mother is 30 years.
3.) 1/2 the weight of the monkey plus the weight of the banana is 1/4 as much the weight of the
total rope and the weight.
4.) The monkey's mother is 1/2 as old as the monkey will be when it is 3 times as old as its mother
was when she was 1/2 as old as the monkey will be when it is as old as its mother will be when she is
4 times as old as the mokey was when it was twice as old as its mother was when she was 1/3 as
old as the monkey ws when it was as old as its mother was when she was three times as old as the
monkey was when it was 1/4 as old as it is now.
5.)How long is the banana?

David's Solution:
While the problem appears to be rather complex, it can be expressed as a system of independent linear equations and therefore can be solved using the principles of Linear Algebra.

From part 1 of the problem statement we yield the following equations.
     Wrope =(1/3) * Lrope (lbs/ft)                                   (eq1)
     Wmonkey = Wweight                                               (eq2)
From part 2 of the problem statement we yield the following equations.
     Wbanana = (2/16) * Lbanana (lbs / inch)          (eq3)
     Lrope = AgeMonkey (feet/years)                         (eq4)
     Wmonkey = (1/16) * AgeMother (lbs / years)    (eq5)
     AgeMonkey + AgeMother = 30 years                 (eq6)
From part 3 of the problem statement we yield the following equations.
     (1/2)*Wmonkey + Wbanana = (1/4)*(Wrope + Wweight)    (eq7)
From part 4 of the problem statement we yield the following equations.
     AgeMother = (1/2)*3*(1/2)*4*2*(1/3)*3*(1/4)*AgeMonkey
                             = (3/2) * AgeMonkey                                                 (eq8)
     You're trying to confuse us with this one

Now I will generate a matrix equation of the form Ax = b.  It is shown that the solution to this equation is x = inv(A)*b.

Equations 1 through 8 can be arranged into the following matrix

[-2/16 1    0     0   0    0    0     0][ Lbanana   ]    [ 0 ]
[    0 0    0     0   1    0    0 -1/16][ Wbanana   ]    [ 0 ]
[    0 0    1     0   0 -1/3    0     0][ Wrope     ]    [ 0 ]
[    0 0    0    -1   1    0    0     0][ Wweight   ]  = [ 0 ]
[    0 1 -1/4  -1/4 1/2    0    0     0][ Wmonkey   ]    [ 0 ]
[    0 0    0     0   0    1   -1     0][ Lrope     ]    [ 0 ]
[    0 0    0     0   0    0 -3/2     1][ AgeMonkey ]    [ 0 ]
[    0 0    0     0   0    0    1     1][ AgeMother ]    [ 30]

The solution of the above system of equations is a tedious exercise intended for simple minds.  Since my mind must be free to imagine and explore I just entered the equations into a numerical solver (MatLab) and received the following answer :

[ Lbanana   ]    [ 5.75  inches]
[ Wbanana   ]    [ 11.5  ounces]
[ Wrope     ]    [ 4     pounds]
[ Wweight   ]  = [ 1.125 pounds]
[ Wmonkey   ]    [ 1.125 pounds]
[ Lrope     ]    [ 12    feet  ]
[ AgeMonkey ]    [ 12    years ]
[ AgeMother ]    [ 18    years ]

Thus the answer to the problem is that the banana is 5.75 inches long.