Maximum value of A1/A2

$\dfrac{A_1}{A_2}=a^2b^2$ . Now, Using AM $\geq$ GM, we have $1 \geq a+2b \geq 2\sqrt{a.2b} \Rightarrow \sqrt{ab} \leq \dfrac{1}{2\sqrt2} \Rightarrow a^2b^2 \leq \dfrac{1}{64}$

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

Let $a+2b=k$ , where $0 < k \le 1$ , $\implies b = \frac 12 (k-a)$ and let $f(a) = \dfrac {A_1}{A_2}$ . Then, we have:

$\begin{aligned} f(a) & = a^2b^2 \\ f'(a) & = 2ab^2 + 2a^2b\frac {db}{da} \\ & = 2a \cdot \frac 1{2^2} (k-a)^2 + 2a^2 \cdot \frac 12 (k-a) \cdot \left(-\frac 12\right) \\ & = \frac {a(k-a)(k-2a)}2 \end{aligned}$

$f'(a) = 0$ , when $a=k$ and $a=\dfrac k2$ . We note that $f''(a) = \dfrac {6a^2-6ka+k^2}2$ , and $f''(k) = \frac {k^2}2 > 0$ , therefore, $f(k)$ is a minimum; $f''\left(\frac k2 \right) = - \frac {k^2}4 < 0$ , therefore, $f\left(\frac k2 \right)$ is a maximum.

$\begin{aligned} f\left(\frac k2 \right) & = \left(\frac k2 \right)^2\left(\frac {k-\frac k2}2 \right)^2 = \left(\frac k2 \right)^2\left(\frac k4 \right)^2 = \frac {k^2}{64} \le \boxed{\dfrac 1{64}} \end{aligned}$