Difference between revisions of "1998 AIME Problems/Problem 14"
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== Problem == | == Problem == | ||
− | An <math>m\times n\times p</math> rectangular box has half the volume of an <math> | + | An <math>m\times n\times p</math> rectangular box has half the volume of an <math>(m + 2)\times(n + 2)\times(p + 2)</math> rectangular box, where <math>m, n,</math> and <math>p</math> are integers, and <math>m\le n\le p.</math> What is the largest possible value of <math>p</math>? |
− | == Solution == | + | == Solution 1 == |
− | < | + | <cmath>2mnp = (m+2)(n+2)(p+2)</cmath> |
Let’s solve for <math>p</math>: | Let’s solve for <math>p</math>: | ||
− | < | + | <cmath>(2mn)p = p(m+2)(n+2) + 2(m+2)(n+2)</cmath> |
− | < | + | <cmath>[2mn - (m+2)(n+2)]p = 2(m+2)(n+2)</cmath> |
− | < | + | <cmath>p = \frac{2(m+2)(n+2)}{mn - 2n - 2m - 4} = \frac{2(m+2)(n+2)}{(m-2)(n-2) - 8}</cmath> |
− | + | Clearly, we want to minimize the denominator, so we test <math>(m-2)(n-2) - 8 = 1 \Longrightarrow (m-2)(n-2) = 9</math>. The possible pairs of factors of <math>9</math> are <math>(1,9)(3,3)</math>. These give <math>m = 3, n = 11</math> and <math>m = 5, n = 5</math> respectively. Substituting into the numerator, we see that the first pair gives <math>130</math>, while the second pair gives <math>98</math>. We now check that <math>130</math> is optimal, setting <math>a=m-2</math>, <math>b=n-2</math> in order to simplify calculations. Since | |
+ | <cmath>0 \le (a-1)(b-1) \implies a+b \le ab+1</cmath> | ||
+ | We have | ||
+ | <cmath>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</cmath> | ||
+ | Where we see <math>(m,n)=(3,11)</math> gives us our maximum value of <math>\boxed{130}</math>. | ||
− | < | + | *Note that <math>0 \le (a-1)(b-1)</math> assumes <math>m,n \ge 3</math>, but this is clear as <math>\frac{2m}{m+2} = \frac{(n+2)(p+2)}{np} > 1</math> and similarly for <math>n</math>. |
− | + | == Solution 2 == | |
+ | |||
+ | Similarly as above, we solve for <math>p,</math> but we express the denominator differently: | ||
+ | |||
+ | <cmath>p=\dfrac{2(m+2)(n+2)}{(m+2)(n+2)-4(m+n+2)} \implies \dfrac{1}{p}=\dfrac{1}{2}-\dfrac{2(m+n+2)}{(m+2)(n+2)}.</cmath> | ||
+ | Hence, it suffices to maximize <math>\dfrac{m+n+2}{(m+2)(n+2)},</math> under the conditions that <math>p</math> is a positive integer. | ||
+ | |||
+ | Then since <math>\dfrac{m+n+2}{(m+2)(n+2)}>\dfrac{1}{2}</math> for <math>m=1,2,</math> we fix <math>m=3.</math> | ||
+ | <cmath>\implies \dfrac{1}{p}=\dfrac{1}{2}-\dfrac{2(n+5)}{5(n+2)}=\dfrac{n-10}{10(n+2)},</cmath> | ||
+ | where we simply let <math>n=11</math> to achieve <math>p=\boxed{130}.</math> | ||
+ | |||
+ | ~Generic_Username | ||
== See also == | == See also == | ||
Line 21: | Line 36: | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 21:05, 29 May 2023
Contents
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 1
Let’s solve for :
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
We have
Where we see
gives us our maximum value of
.
- Note that
assumes
, but this is clear as
and similarly for
.
Solution 2
Similarly as above, we solve for but we express the denominator differently:
Hence, it suffices to maximize
under the conditions that
is a positive integer.
Then since for
we fix
where we simply let
to achieve
~Generic_Username
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.