A) \[\frac{{{c}^{3}}}{2ab}\]
B) \[\frac{{{c}^{3}}}{\sqrt{2ab}}\]
C) \[\frac{{{c}^{3}}}{ab}\]
D) \[\frac{{{c}^{3}}}{\sqrt{ab}}\]
Correct Answer: B
Solution :
Given, \[{{a}^{2}}{{x}^{4}}+{{b}^{2}}{{y}^{4}}={{c}^{6}}\] \[\Rightarrow \] \[{{y}^{4}}=\frac{{{c}^{6}}-{{a}^{2}}{{x}^{4}}}{{{b}^{2}}}\] \[\Rightarrow \] \[y={{\left( \frac{{{c}^{6}}-{{a}^{2}}{{x}^{4}}}{{{b}^{2}}} \right)}^{1/4}}\] Let \[z=xy=x{{\left( \frac{{{c}^{6}}-{{a}^{2}}{{x}^{4}}}{{{b}^{2}}} \right)}^{1/4}}\] \[\Rightarrow \] \[z={{\left( \frac{{{c}^{6}}{{x}^{4}}-{{a}^{2}}{{x}^{6}}}{{{b}^{2}}} \right)}^{\frac{1}{4}}}\] \[\Rightarrow \]\[\frac{dz}{dx}=\frac{1}{4}{{\left( \frac{{{c}^{6}}{{x}^{4}}-{{a}^{2}}{{x}^{8}}}{{{b}^{2}}} \right)}^{-3/4}}.\left( \frac{4{{x}^{3}}{{c}^{6}}}{{{b}^{2}}}-\frac{8{{a}^{2}}{{x}^{7}}}{{{b}^{2}}} \right)\] For maxima and minima, \[\frac{dy}{dx}=0\] \[\Rightarrow \] \[\frac{1}{4}{{\left( \frac{{{c}^{6}}{{x}^{4}}-{{a}^{2}}{{x}^{8}}}{{{b}^{2}}} \right)}^{-3/4}}\times \left( \frac{4{{x}^{3}}{{c}^{6}}}{{{b}^{2}}}-\frac{8{{a}^{2}}{{x}^{7}}}{{{b}^{2}}} \right)=0\] \[\Rightarrow \] \[\frac{4{{x}^{3}}{{c}^{6}}}{{{b}^{2}}}-\frac{8{{a}^{2}}{{x}^{7}}}{{{b}^{2}}}=0\] \[\Rightarrow \] \[\frac{4{{x}^{3}}}{{{b}^{2}}}({{c}^{6}}-2{{a}^{2}}{{x}^{4}})=0\] \[\Rightarrow \] \[{{c}^{6}}-2{{a}^{2}}{{x}^{4}}=0\] \[\Rightarrow \] \[{{x}^{4}}=\frac{{{c}^{6}}}{2{{a}^{2}}}\] \[\Rightarrow \] \[x=\pm \frac{{{c}^{6/4}}}{{{2}^{1/4}}{{a}^{2/4}}}=\pm \frac{{{c}^{3/2}}}{{{2}^{1/4}}.{{a}^{1/2}}}\] \[\therefore \]At \[x=\pm \frac{{{c}^{3/2}}}{{{2}^{1/4}}\sqrt{a}},z\]is maximum \[\therefore \] \[z={{\left[ \frac{{{c}^{6}}}{{{b}^{2}}}\left( \frac{{{c}^{6}}}{2{{a}^{2}}} \right)-\frac{{{a}^{2}}}{{{b}^{2}}}{{\left( \frac{{{c}^{6}}}{2{{a}^{2}}} \right)}^{2}} \right]}^{1/4}}\] \[={{\left[ \frac{{{c}^{12}}}{2{{a}^{2}}{{b}^{2}}}.\frac{{{c}^{12}}}{4{{a}^{2}}{{b}^{2}}} \right]}^{1/4}}\] \[={{\left( \frac{{{c}^{12}}}{4{{a}^{2}}{{b}^{2}}} \right)}^{1/4}}\] \[=\frac{{{c}^{3}}}{\sqrt{2ab}}\]
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