Expert: Josh - 3/22/2004

Question
four statues stand on rectangular blocks with square tops,all of different sizes.For each block,its volume in cubic feet (less than 5oo) is equal to the total area of its six faces in square feet.One of the blocks is a 6 foot cube (volume 6 times 6 times 6= 216 cu.ft and area 6 faces each with an area 36 sq.ft= 216 sq ft) give the dimensions,all whole numbers of feet,of any of the other blocks (2 are lower than the cube and one higher)

Answer
Hi David,

First, set up this problem algebraically.
Let h(i) represent the height of the ith block.
Similarly, let w(i) and d(i) denote the width and depth of the ith block; where i=1,2,3,4,5,6.

Givens:
For the first block i=1,
h(1)=w(1)=d(1)=6.

One block higher than block 1, ie.,
h(2)>h(1). ...[Fact 1]

Two blocks lower than block 1, ie.,
h(3),h(4)<h(1).   ...[Fact 2]

Extension: Assume that two other blocks have the same height as block 1, ie.,
h(5)=h(6)=h(1) ...[Fact 3]

Volume(i)=h(i)*w(i)*d(i)
Surface_Area(i)=2*[w(i)*d(i)+h(i)*w(i)+h(i)*d(i)]
Constraint: Volume(i)=Surface_Area(i) for all blocks, i.

h*w*d=2*[w*d+h*w+h*d] ...[General Constraint, G1]

If w=d, ...[Assumption, A1]
h*w^2=2*[w^2+2*h*w]
we obtain the quadratic equation,
(h-2)*w^2-4*h*w=0
w[(h-2)*w-4*h]=0
Solving for w, assuming w not equal to 0,
  w=4h/(h-2) ...[Condition, #1]
will always satisfy the constraint.

Quick check: We were told that h(1)=6,w(1)=6,d(1)=6 is one feasible solution, check if this satisfies the condition [#1].
Substituting h=6 into [#1], we get
w=4*6/(6-2)=24/4=6.
It follows that d=w=6 under our earlier assumption [A1].

Using [#1], we can easily obtain other possible solutions.

(a) Why not let height of block 2, h(2)=7,
Then, w(2)=4*h(2)/(h(2)-2)=4*7/(7-2)=28/5.
Unfortunately, this does not divide nicely.
So, try again, when h(2)=10, w(2)=4*10/(10-2)=5
Therefore, [h(2),w(2),d(2)]=[10,5,5].

(b) Next, try lower heights for block 3.
Repeat what we did in part (a).
Notice that when h(3)=3, w=4*3/(3-2)=12 is an integer.
So, [h(3),w(3),d(3)]=[3,12,12].
Observe that Surface_Area(3)=2[3*6+6*6+6*3]=432 and Volume(3)=6*12*12=432.

(c) By inspection, [h(4),w(4),d(4)]=[4,8,8] also works.

For block 5 and block 6, we derive things differently.

Let h(5)=h(6)=h(1)=6.
Assume that width, "w", is not equal to depth ,"d".
Revisiting the general constraint [G1], repeated below,
h*w*d=2*[w*d+h*w+h*d]
We have,
6wd=2[wd+6(w+d)], expanding this expression
6wd=2wd+12w+12d , canceling common factors, we get
dw=3(d+w) ...[Constraint, #2]

Which means that the ratio between the product of "width" and "depth", AND the sum of "width" and "depth" must be 3.

Does this make sense?
Well, check that it works for the case where d=6.
6w=3(6+w) has solution w=6.
Which leads to the prior example of 6,6,6.

Now, try something different.
Using a program that I have written, I searched the following combinations where the dimension is between 1 and 100.

for(d=1, d<100, d=d+1)
for(w=1, w<100, w=w+1)
 if (((d*w)/(d+w))==3)
  printf('d=%d,w=%d',d,w)

answer =
d=4,w=12
d=6,w=6
d=12,w=4

So, [h(5),w(5),d(5)]=[6,4,12].
[h(6),w(6),d(6)]=[6,12,4].

So, there you have it. The solutions are easily found with the help of a computer program. So long as you formulate the problem properly.

Block 1 (given)
 [h(1),w(1),d(1)]=[6,6,6]
Block 2 is higher than Block 1
 [h(2),w(2),d(2)]=[10,5,5].
Block 3 & 4 are lower than Block 1 in height.
 [h(3),w(3),d(3)]=[3,12,12].
 [h(4),w(4),d(4)]=[4,8,8]
Block 5 & 6 are equal in height.
 [h(5),w(5),d(5)]=[6,4,12].
 [h(6),w(6),d(6)]=[6,12,4].

Cheers,
Josh

P.S. Do I get to share the prize? Let's split it 50:50