Explanation

(c)
Given, $a^{2}x^{4} + b^{2}y^{4} = c^{6}$
$\Rightarrow \ \ \ y = \left( \frac{c^{6} - a^{2}x^{4}}{b^{2}} \right)^{1/4}$
and let $f(x) = xy = x\left( \frac{c^{6} - a^{2}x^{4}}{b^{2}} \right)^{1/4}$
$\Rightarrow \ \ f(x) = \left( \frac{c^{6}x^{4} - a^{2}x^{8}}{b^{2}} \right)^{1/4}$
On differentiating w.r.t. $x$, we get
$f^{'}(x) = \frac{1}{4}\left( \frac{c^{6}x^{4} - a^{2}x^{8}}{b^{2}} \right)^{1/4}\left( \frac{{4x}^{3}c^{6}}{b^{2}} - \frac{8x^{7}a^{2}}{b^{2}} \right)$
For maxima or minimum, put $f^{'}(x) = 0$
$\Rightarrow \ \ \frac{4x^{3}c^{6}}{b^{2}} = \frac{8x^{7}a^{2}}{b^{2}} = 0$
$\Rightarrow \ \frac{4x^{3}}{b^{2}}\left( c^{6} - 2a^{2}x^{4} \right) = 0$
$\Rightarrow \ \ x^{4} = \frac{c^{6}}{2a^{2}}\ \Rightarrow \ \ \ \pm \frac{c^{3/2}}{c^{1/4}\sqrt{a}}\ \$
At $x = \frac{c^{3/2}}{2^{1/4}\sqrt{a}},\ f(x)$ will be maximum
$\therefore\ \ f\left( \frac{c^{3/2}}{2^{1/4}\sqrt{a}} \right) = \left( \frac{c^{12}}{2a^{2}b^{2}} - \frac{c^{12}}{4a^{2}b^{2}} \right)^{1/4}$
$= \left( \frac{c^{12}}{4a^{2}b^{2}} \right)^{1/4} = \frac{c^{3}}{\sqrt{2ab}}$