Q108 / 160 · AP EAMCET 2024 (21 May Shift 2) · Mathematics

If $a^2 x^4+b^2 y^4=c^6$ then maximum value of $x y=$

  • A
    $\frac{c^3}{2 a b}$
  • Correct answer
    B
    $\frac{c^3}{\sqrt{2 a b}}$
  • C
    $\frac{c^3}{a b}$
  • D
    $\frac{c^3}{\sqrt{a b}}$

Solution

$\begin{aligned} & a^2 x^4+b^2 y^4=c^6 \\ & \Rightarrow \frac{a^2 x^4+b^2 y^4}{2} \geq \sqrt{a^2 b^2 x^4 y^4} \Rightarrow \frac{c^6}{2} \geq a b x^2 y^2 \\ & \Rightarrow x^2 y^2 \leq \frac{c^6}{2 a b} \Rightarrow x y \leq \frac{c^3}{\sqrt{2 a b}} .\end{aligned}$

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

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