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Q. If $ {{a}^{2}}{{x}^{4}}+{{b}^{2}}{{y}^{4}}={{c}^{6}}, $ then maximum value of $ xy $ is

Rajasthan PETRajasthan PET 2001

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}} $