The maximum value of $x^4 y^4$ given the constraint $a^2 x^4 + b^2 y^4 = c^6$ is

  • A
    $\frac{c^{12}}{16 a^2 b^2}$
  • B
    $\frac{c^{12}}{4 a^2 b^2}$
  • C
    $\frac{c^{12}}{8 a^2 b^2}$
  • D
    $\frac{c^{12}}{2 a^2 b^2}$

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The sum of two non-zero numbers is $4$. The minimum value of the sum of their reciprocals is

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