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/8/b/1/8b1c33fbb4e00bb1281b037010052fe6d01b6f01.png)
Let’s solve for :
![$(2mn)p = p(m+2)(n+2) + 2(m+2)(n+2)$](http://latex.artofproblemsolving.com/d/5/f/d5ff63a88caf1ac4552d9215366e18fab5e144d8.png)
![$p = \frac{2(m+2)(n+2)}{mn - 2n - 2m - 4}$](http://latex.artofproblemsolving.com/f/1/0/f108f6bdc2234b440849dafc9bf269e8e77d0e8d.png)
For the denominator, we will use a factoring trick (colloquially known as SFFT), which states that .
![$p = \frac{2(m+2)(n+2)}{(m-2)(n-2) - 8}$](http://latex.artofproblemsolving.com/b/8/4/b84a61a6b6fdd06d3924cbb85b37bbabef83703f.png)
Clearly, we want to minimize the denominator, so . 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 can quickly test for the denominator assuming other values to convince ourselves that
is the best possible value for the denominator. Hence, the solution is
.
Proof that setting the denominator to
is optimal: Suppose
, and suppose for the sake of contradiction that there exist
such that
for some
and such that
This implies that
and
Substituting gives
which we rewrite as
Next, note that for
to be positive, we must have
and
be positive, so
So
//shouldn't this be 24 + 4(9+d)? then the entire proof is fallacious (it results in 2(1 + 3 + 3) < 15 which is true)
Next, we must have that
and
are positive, so
and
. Also,
by how we defined
. So
a contradiction. We already showed above that there are some values of
and
such that
that work, so this proves that one of these pairs of values of
and
must yield the maximal value of
.
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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