Q21 / 58 · KVPY 2016 (SA) · Mathematics

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

  • A
    $\frac{1}{16}$
  • Correct answer
    B
    $\frac{1}{64}$
  • C
    $\frac{1}{16 \sqrt{2}}$
  • D
    $\frac{1}{32}$

Solution

$a+2 b \leq 1$ $A_{1}=\pi a^{2} b^{6}$ $A_{2}=\pi b^{4}$ $\Rightarrow \frac{A_{1}}{A_{2}}=a^{2} b^{2}$ $\& 1 \geq a+2 b \geq 2 \sqrt{2 a b}(A M \geq G M)$ $\Rightarrow 1 \geq 2 \sqrt{2 a b}$ $\Rightarrow 1 \geq 4-2 a b$ $\Rightarrow \frac{1}{64} \geq a^{2} b^{2}$

Hence, the correct answer is option: 'B'.

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