Mock AIME 1 2010 Problems/Problem 10
Problem
Find the last three digits of the largest possible value of ![\[\frac{a^2 b^6}{a^{2 \log_2 a} (a^2 b)^{\log_2 b}},\]](http://latex.artofproblemsolving.com/2/7/b/27b1df5754d36aaa8dd409431106a5e436b6dd01.png)
where 
and 
are positive reals.
Solution 1
Let 
and 
and 
.
To help find an inequality we can use to optimize this expression, we combine all of the terms into a single exponential function: \begin{align*} \frac{a^2 b^6}{a^{2 \log_2 a} (a^2 b)^{\log_2 b}} &= \frac{a^2b^6}{a^{2A-2B}b^B} \\ &= a^{2(1-A-B)}b^{6-B} \\ &= 2^{2A(1-A-B)}2^{B(6-B)} \\ &= 2^{2A-2A^2-2AB+6B-B^2} \end{align*}
Now, because 
is monotone increasing, we desire to maximize 
Because this is a parabola facing downwards (if we treat one of 
and 
to be constant while the other varies), we can use the fact that the vertex of a parabola of the form 
occurs at 
(one can prove this by looking at the quadratic formula or with derivatives).
Now, we will optimize and in terms of the other if the other variable is constant.
Treating to be constant, we have the parabola 
, which has a vertex (and thereby a maximum) at 
Treating to be constant, we have the parabola 
, which has a vertex (and thereby a maximum) at 
Now, we have a system of two equations: \begin{align*} B&=3-A \\ 2A&=1-B \implies B = 1-2A \\ \implies 1-2A &= 3-A \\ A &= -2 \\ \implies B &= 3-(-2)=5 \end{align*}
Because logarithms have a range of all real numbers, these values of and are attainable.
Now from the above optimization, we know that 
, and so the given expression, which equals 
, is at most equal to 
. This value is attained when 
and 
(by definition of and ), and so this maximum is possible. Thus, our answer is 
.
See Also
| Mock AIME 1 2010 (Problems, Source) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
