Question Number 174227 by mnjuly1970 last updated on 27/Jul/22
$$ \\ $$$$\:\:\:{Q}\:: \\ $$$$\:\:\:\:\:\:\:{Max}_{\underset{{x}>\mathrm{1}} {\:}} \:\left(\frac{\:{x}^{\:\mathrm{4}} −{x}^{\:\mathrm{2}} }{{x}^{\:\mathrm{6}} +\:\mathrm{2}{x}^{\:\mathrm{3}} −\:\mathrm{1}}\:\right)\:=\:? \\ $$$$ \\ $$
Commented by infinityaction last updated on 27/Jul/22
$$\:\:\:\:\frac{\mathrm{1}}{\mathrm{6}}\:??? \\ $$
Commented by mnjuly1970 last updated on 27/Jul/22
$${yes}\:\:{Sir}… \\ $$
Commented by infinityaction last updated on 27/Jul/22
$$\:\:{p}\:=\:\:\:\frac{{x}−\frac{\mathrm{1}}{{x}}}{{x}^{\mathrm{3}} −\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+\mathrm{2}}\:=\:\frac{{x}−\frac{\mathrm{1}}{{x}}}{\left({x}−\frac{\mathrm{1}}{{x}}\right)\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\mathrm{1}\right)+\mathrm{2}} \\ $$$$\:\:\:\:\:{p}\:\:\:=\:\:\frac{{x}−\frac{\mathrm{1}}{{x}}}{\left({x}−\frac{\mathrm{1}}{{x}}\right)\left\{\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\mathrm{3}\right\}+\mathrm{2}} \\ $$$$\:\:{let}\:{x}−\frac{\mathrm{1}}{{x}}\:=\:{g} \\ $$$$\:\:\:\:{p}\:\:=\:\:\frac{{g}}{{g}\left({g}^{\mathrm{2}} +\mathrm{3}\right)+\mathrm{2}}\:\Rightarrow\frac{\mathrm{1}}{{g}^{\mathrm{2}} +\frac{\mathrm{2}}{{g}}+\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:{p}\:\:=\:\:\frac{\mathrm{1}}{{g}^{\mathrm{2}} +\frac{\mathrm{1}}{{g}}+\frac{\mathrm{1}}{{g}}+\mathrm{3}} \\ $$$$\:\:\:{p}_{{max}\:\:} =\:\:\frac{\mathrm{1}}{\mathrm{3}+\mathrm{3}}\:\:=\:\frac{\mathrm{1}}{\mathrm{6}} \\ $$
Commented by mnjuly1970 last updated on 27/Jul/22
$$\:\:{very}\:{nice}\:{solution}\: \\ $$$${thanks}\:{alot}\:{sir} \\ $$