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If $a^{2} x^{4}+b^{2} y^{4}=c^{4}$, then the maximum value of xy is
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Verified Answer
The correct answer is:
$\frac{c^{2}}{\sqrt{2 a b}}$
If the sum of two positive quantities is a constant, then their product is maximum, when they are equal. $\therefore \mathrm{a}^{2} \mathrm{x}^{4} \cdot \mathrm{b}^{2} \mathrm{y}^{2}$ is maximum when
$$
a^{2} x^{4}=b^{2} y^{4}=\frac{1}{2}\left(a^{2} x^{4}=\frac{1}{2}\left(a^{2} x^{4}+b^{2} y^{4}\right)=\frac{c^{4}}{2}\right.
$$
$\therefore$ maximum value of $a^{2} x^{4} \cdot b^{2} y^{4}=\frac{c^{4}}{2} \cdot \frac{c^{4}}{2}=\frac{c^{8}}{4}$
Maximum value of $x y=\left(\frac{c^{8}}{4 a^{2} b^{2}}\right)^{1 / 4}=\frac{c^{2}}{\sqrt{2 a b}}$
$$
a^{2} x^{4}=b^{2} y^{4}=\frac{1}{2}\left(a^{2} x^{4}=\frac{1}{2}\left(a^{2} x^{4}+b^{2} y^{4}\right)=\frac{c^{4}}{2}\right.
$$
$\therefore$ maximum value of $a^{2} x^{4} \cdot b^{2} y^{4}=\frac{c^{4}}{2} \cdot \frac{c^{4}}{2}=\frac{c^{8}}{4}$
Maximum value of $x y=\left(\frac{c^{8}}{4 a^{2} b^{2}}\right)^{1 / 4}=\frac{c^{2}}{\sqrt{2 a b}}$
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