Answers 309passes 2 times along length, 4 times along breadth, and 6 times along depth.
Therefore 4 feet divided by 2, 4, and 6 will give us 2 feet, 1 foot, and "'3 foot
respectively for the length, breadth, and depth of the largest possible parcel.
The following general solution is by Mr. Alexander Fraser. Let the string
pass a times along length x, b times along breadth y, and c times along depth
z, and let length of string be m.
Then ax + by + cz = m. Find maximum value of xyz.
First find maximum area of xy.
Put ax + by _ _ n, x _ _ --, n - by xy _ -_ n y - -b y,^2 dxy dy _ _ - - -n 2b_ y _ 0,
a a a a a. n b n
· .y = 2b' or ~ = 2'
· .. ax also = 1, and ax = by.
Similarly, ax = by = cz = ~.
. m m m m^3
·. x = 3a'y = 3b' z =)c' and xyz = 27abc'
In the case of the puzzle, a = 2, b = 4, c = 6, m = 12.
. '. x = 2,y = I, z = 13.
xyz = IY.!.
314. THE STONE PEDESTAL
The cube of a square number is always a square. Thus:-
The cube of 1 is I, the square of l.
The cube of 4 is 64, the square of 8.
The cube of 9 is 729, the square of 27.
The cube of 16 is 4,096, the square of 64.And so on.
We were told to look at the illustration. If there were one block in pedestal
and one in base, the base would be entirely covered, which it was not. If 64
in pedestal and base, the side of the former would measure 4 feet, and the side
of square 8 feet. A glance will show that this is wrong. But 729 blocks in each
case is quite in agreement with the illustration, for the width of the pedestal