KVPY 2016 SA Question 1

$\begin{aligned} \frac {A_1}{A_2} & = \frac {\pi \left(ab^3\right)^2}{\pi \left(b^2\right)^2} = \frac {a^2b^6}{b^4} = a^2b^2 \end{aligned}$

Now from $a + 2b \le 1$ , using AM-GM inequality :

$\begin{aligned} 1 \ge a + 2b & \ge 2\sqrt{2ab} \\ \implies 2\sqrt{2ab} & \le 1 & \small \color{#3D99F6} \text{Raising both sides to the power of 4} \\ 64a^2b^2 & \le 1 \\ a^2b^2 & \le \frac 1{64} \\ \implies \frac {A_1}{A_2} & \le \boxed{\dfrac 1{64}} \end{aligned}$