1998 AIME Problems/Problem 14
Problem
An 
rectangular box has half the volume of an 
rectangular box, where 
and 
are integers, and 
What is the largest possible value of ?
Solution
![\[2mnp = (m+2)(n+2)(p+2)\]](http://latex.artofproblemsolving.com/d/4/8/d488161a60c3b671e6a099deae2b1bd8743f01e9.png)
Let’s solve for :
![\[(2mn)p = p(m+2)(n+2) + 2(m+2)(n+2)\]](http://latex.artofproblemsolving.com/7/b/1/7b143cfbb2985678c4a7d6424e032e016fa2a879.png)
![\[[2mn - (m+2)(n+2)]p = 2(m+2)(n+2)\]](http://latex.artofproblemsolving.com/8/d/8/8d85d749a6a153c00912c02d1cc6bd841fb8f0f2.png)
![\[p = \frac{2(m+2)(n+2)}{mn - 2n - 2m - 4} = \frac{2(m+2)(n+2)}{(m-2)(n-2) - 8}\]](http://latex.artofproblemsolving.com/4/c/a/4ca431090d2a4c440e06936242d05d118a9b4789.png)
Clearly, we want to minimize the denominator, so we test 
. The possible pairs of factors of 
are 
. These give 
and 
respectively. Substituting into the numerator, we see that the first pair gives 
, while the second pair gives 
. We now check that is optimal, setting 
, 
in order to simplify calculations. Since
![\[0 \le (a-1)(b-1) \implies a+b \le ab+1\]](http://latex.artofproblemsolving.com/f/c/b/fcbb8bd5842ec9e3e12154c710e3221711c4ec51.png)
We have
![\[p = \frac{2(a+4)(b+4)}{ab-8} = \frac{2ab+8(a+b)+32}{ab-8} \le \frac{2ab+8(ab+1)+32}{ab-8} = 10 + \frac{120}{ab-8} \le 130\]](http://latex.artofproblemsolving.com/b/4/b/b4b5f08e82b91dea4d2a91f26251aa9db9b5a164.png)
Where we see 
gives us our maximum value of 
.
- Note that
assumes 
, but this is clear as 
and similarly for 
.
assumes 
, but this is clear as 
and similarly for 
.See also
| 1998 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
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