If {a^2}{x^4} + {b^2}{y^4} = {c^6},then the m | vedclass

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(C) Given the equation ${a^2}{x^4} + {b^2}{y^4} = {c^6}$.Using the $AM$-$GM$ inequality,we know that for positive numbers $u$ and $v$,$\frac{u+v}{2} \

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$\frac{{{c^3}}}{{\sqrt {2ab} }}$

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(C) Given the equation ${a^2}{x^4} + {b^2}{y^4} = {c^6}$.Using the $AM$-$GM$ inequality,we know that for positive numbers $u$ and $v$,$\frac{u+v}{2} \ge \sqrt{uv}$.Let $u = {a^2}{x^4}$ and $v = {b^2}{y^4}$.Then $\frac{{{a^2}{x^4} + {b^2}{y^4}}}{2} \ge \sqrt{{a^2}{x^4} \cdot {b^2}{y^4}}$.Substituting the given equation: $\frac{{{c^6}}}{2} \ge \sqrt{{a^2}{b^2}{x^4}{y^4}}$.$\frac{{{c^6}}}{2} \ge ab{x^2}{y^2}$.${x^2}{y^2} \le \frac{{{c^6}}}{{2ab}}$.Taking the square root on both sides,we get $xy \le \sqrt{\frac{{{c^6}}}{{2ab}}}$.Therefore,the maximum value of $xy$ is $\frac{{{c^3}}}{{\sqrt {2ab} }}$.

If ${a^2}{x^4} + {b^2}{y^4} = {c^6}$,then the maximum value of $xy$ is

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