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The box (Posted on 2009-01-02) Difficulty: 2 of 5
A box has integer dimensions, and when each one is increased by 2, its volume doubles. What is the largest possible dimension?

  Submitted by pcbouhid    
Rating: 5.0000 (1 votes)
Solution: (Hide)
(a+2)(b+2)(c+2) = 2abc

c(a+2)(b+2) + 2(a+2)(b+2) = 2abc

c[2ab - (a+2)(b+2)] = 2(a+2)(b+2)

c = 2(a+2)(b+2)/[2ab-(a+2)(b+2)]

c = 2(a+2)(b+2)/(ab-2a-2b-4)

c = (2ab + 4a + 4b + 8)/(ab-2a-2b-4)

c = [(2ab-4a-4b-8) + (8a+8b+16)]/(ab-2a-2b-4)

c = 2 + (8a+8b+16)/(ab-2a-2b-4).

(c-2)/8 = (a+b+2)/(ab-2a-2b-4).

Making c'=c-2, b'=b-2, a'=a-2, we have:

c'/8 = (a'+b'+6)/(a'b'-8).

We cannot have a or b = 1 or 2, because (1+2)/1 and (2+2)/2 are already ≥ 2 (so multiplying by (c+2)/c gives > 2). Then a, b ≥3, so a', b' ≥ 1, (a'+b') ≤ (a'b'+1) and (a´+ b´+ 6) ≤ (a´b´+7).

Hence c'/8 ≤ (a'b'+7)/(a'b'-8) = 1 + 15/(a'b'-8) ≤ 1 + 15.

Hence c´ ≤ 128, and finally c ≤ 130.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
Solutionspoiler--all solutionsxdog2009-01-05 00:45:29
re: Simple stupid way?Charlie2009-01-03 12:12:13
QuestionSimple stupid way?elementofsurprize2009-01-03 08:24:22
somewhat analytical solutionDaniel2009-01-02 21:41:48
Some Thoughtscomputer exploration (spoiler)Charlie2009-01-02 18:33:28
re: Solution?Sing4TheDay2009-01-02 16:52:11
Some ThoughtsSolution?Sing4TheDay2009-01-02 16:02:23
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