Let a and b be two positive real numbers such | vedclass

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(B) Given $a, b > 0$ and $a+2b \leq 1$. Radius of circle $C_1 = ab^3$ and radius of circle $C_2 = b^2$. Area $A_1 = \pi(ab^3)^2 = \pi a^2b^6$. Area $A

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$\frac{1}{64}$

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(B) Given $a, b > 0$ and $a+2b \leq 1$. Radius of circle $C_1 = ab^3$ and radius of circle $C_2 = b^2$. Area $A_1 = \pi(ab^3)^2 = \pi a^2b^6$. Area $A_2 = \pi(b^2)^2 = \pi b^4$. Thus,$\frac{A_1}{A_2} = \frac{\pi a^2b^6}{\pi b^4} = a^2b^2 = (ab)^2$. Using the $AM$-$GM$ inequality for $a$ and $2b$: $\frac{a+2b}{2} \geq \sqrt{a \cdot 2b} = \sqrt{2ab}$. Since $a+2b \leq 1$,we have $\frac{1}{2} \geq \sqrt{2ab}$. Squaring both sides,$\frac{1}{4} \geq 2ab$,which implies $ab \leq \frac{1}{8}$. Therefore,$(ab)^2 \leq (\frac{1}{8})^2 = \frac{1}{64}$. The maximum value of $\frac{A_1}{A_2}$ is $\frac{1}{64}$.

Let $a$ and $b$ be two positive real numbers such that $a+2b \leq 1$. Let $A_1$ and $A_2$ be respectively the areas of circles with radii $ab^3$ and $b^2$. Then,the maximum possible value of $\frac{A_1}{A_2}$ is

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