MCQ
If ${a^2}{x^4} + {b^2}{y^4} = {c^6},$ then maximum value of $xy $ is
- A${{{c^2}} \over {\sqrt {ab} }}$
- B${{{c^3}} \over {ab}}$
- ✓${{{c^3}} \over {\sqrt {2ab} }}$
- D${{{c^3}} \over {2ab}}$
Hence $f(x) = xy = x{\left( {\frac{{{c^6} - {a^2}{x^4}}}{{{b^2}}}} \right)^{1/4}}$
==> $f(x) = {\left( {\frac{{{c^6}{x^4} - {a^2}{x^8}}}{{{b^2}}}} \right)^{1/4}}$
Differentiate $f(x)$ with respect to $ x$ , then
$f'(x) = \frac{1}{4}{\left( {\frac{{{c^6}{x^4} - {a^2}{x^8}}}{{{b^2}}}} \right)^{ - 3/4}}\left( {\frac{{4{x^3}{c^6}}}{{{b^2}}} - \frac{{8{x^7}{a^2}}}{{{b^2}}}} \right)$
Put $f'(x) = 0$,
$\frac{{4{x^3}{c^6}}}{{{b^2}}} - \frac{{8{x^7}{a^2}}}{{{b^2}}} = 0$
==> ${x^4} = \frac{{{c^6}}}{{2{a^2}}}$ ==> $x = \pm \frac{{{c^{3/2}}}}{{{2^{1/4}}\sqrt a }}$
At $x = \frac{{{c^{3/2}}}}{{{2^{1/4}}\sqrt a }}$ the $f(x)$ will be maximum, so
The minimum value $f{\rm{ }}\left( {\frac{{{c^{3/2}}}}{{{2^{1/4}}\sqrt a }}} \right)\, = \,{\left( {\frac{{{c^{12}}}}{{2{a^2}{b^2}}} - \frac{{{c^{12}}}}{{4{a^2}{b^2}}}} \right)^{1/4}} $
$= {\left( {\frac{{{c^{12}}}}{{4{a^2}{b^2}}}} \right)^{1/4}} = \frac{{{c^3}}}{{\sqrt {2ab} }}$.
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