A) \[\frac{{{c}^{2}}}{\sqrt{ab}}\]
B) \[\frac{{{c}^{3}}}{ab}\]
C) \[\frac{{{c}^{3}}}{\sqrt{2ab}}\]
D) \[\frac{{{c}^{3}}}{2ab}\]
Correct Answer: C
Solution :
\[{{a}^{2}}{{x}^{4}}+{{b}^{2}}{{y}^{4}}={{c}^{6}}\] Þ \[y={{\left( \frac{{{c}^{6}}-{{a}^{2}}{{x}^{4}}}{{{b}^{2}}} \right)}^{1/4}}\] Hence \[f(x)=xy=x{{\left( \frac{{{c}^{6}}-{{a}^{2}}{{x}^{4}}}{{{b}^{2}}} \right)}^{1/4}}\] Þ \[f(x)={{\left( \frac{{{c}^{6}}{{x}^{4}}-{{a}^{2}}{{x}^{8}}}{{{b}^{2}}} \right)}^{1/4}}\] 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 \[{f}'(x)=0\], \[\frac{4{{x}^{3}}{{c}^{6}}}{{{b}^{2}}}-\frac{8{{x}^{7}}{{a}^{2}}}{{{b}^{2}}}=0\] Þ \[{{x}^{4}}=\frac{{{c}^{6}}}{2{{a}^{2}}}\] Þ \[x=\pm \frac{{{c}^{3/2}}}{{{2}^{1/4}}\sqrt{a}}\] At \[x=\frac{{{c}^{3/2}}}{{{2}^{1/4}}\sqrt{a}}\] the \[f(x)\] will be maximum, so \[f\text{ }\left( \frac{{{c}^{3/2}}}{{{2}^{1/4}}\sqrt{a}} \right)\,=\,{{\left( \frac{{{c}^{12}}}{2{{a}^{2}}{{b}^{2}}}-\frac{{{c}^{12}}}{4{{a}^{2}}{{b}^{2}}} \right)}^{1/4}}={{\left( \frac{{{c}^{12}}}{4{{a}^{2}}{{b}^{2}}} \right)}^{1/4}}=\frac{{{c}^{3}}}{\sqrt{2ab}}\].
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