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Mock AIME 1 2010 Problems/Problem 10

Problem

Find the last three digits of the largest possible value of \[\frac{a^2 b^6}{a^{2 \log_2 a} (a^2 b)^{\log_2 b}},\] where $a$ and $b$ are positive reals.

Solution 1

Let $A=\log_2a$ and $B=\log_2b$ $\implies a=2^A$ and $b=2^B$.

To help find an inequality we can use to optimize this expression, we combine all of the terms into a single exponential function: a2b6a2log2⁡a(a2b)log2⁡b=a2b6a2A−2BbB=a2(1−A−B)b6−B=22A(1−A−B)2B(6−B)=22A−2A2−2AB+6B−B2

Now, because $2^x$ is monotone increasing, we desire to maximize $2A-2A^2-2AB+6B-B^2.$ Because this is a parabola facing downwards (if we treat one of $A$ and $B$ to be constant while the other varies), we can use the fact that the vertex of a parabola of the form $y=ax^2+bx+c$ occurs at $x=\frac{-b}{2a}$ (one can prove this by looking at the quadratic formula or with derivatives).

Now, we will optimize $A$ and $B$ in terms of the other if the other variable is constant.

Treating $A$ to be constant, we have the parabola $f(B)=-B^2+(6-2A)B-2A^2+2A$, which has a vertex (and thereby a maximum) at $B=\frac{2A-6}{-2}=3-A.$

Treating $B$ to be constant, we have the parabola $f(A)=-2A^2+(2-2B)A+6B-B^2$, which has a vertex (and thereby a maximum) at $A=\frac{2B-2}{-4}=\frac{1-B}{2}.$

Now, we have a system of two equations: B=3−A2A=1−B⟹B=1−2A⟹1−2A=3−AA=−2⟹B=3−(−2)=5

Because logarithms have a range of all real numbers, these values of $A$ and $B$ are attainable.

Now from the above optimization, we know that $2A-2A^2-2AB+6B-B^2\leq 2(-2)-2(-2)^2-2(-2)(5)+6(5)-5^2=13$, and so the given expression, which equals $2^{2A-2A^2-2AB+6B-B^2}$, is at most equal to $2^{13}=8192$. This value is attained when $a=2^{-2}$ and $b=2^5$ (by definition of $A$ and $B$), and so this maximum is possible. Thus, our answer is $\boxed{192}$.

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
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