Question Number 131965 by liberty last updated on 10/Feb/21
Commented by liberty last updated on 10/Feb/21
Commented by TheSupreme last updated on 10/Feb/21
$${T}={t}_{\mathrm{1}} +{t}_{\mathrm{2}} \\ $$$$\left.\mathrm{1}\right)\:\:\sqrt{\mathrm{3}^{\mathrm{2}} +{x}^{\mathrm{2}} }=\mathrm{6}{t}_{\mathrm{1}} \:\: \\ $$$$\left.\mathrm{2}\right)\:\mathrm{8}−{x}=\mathrm{8}{t}_{\mathrm{2}} \\ $$$${T}=\frac{\sqrt{\mathrm{9}+{x}^{\mathrm{2}} }}{\mathrm{6}}+\mathrm{1}−\frac{{x}}{\mathrm{8}} \\ $$$$\mathrm{0}<{x}<\mathrm{8} \\ $$$$\frac{{dT}}{{dx}}=\:\frac{{x}}{\:\mathrm{6}\sqrt{\mathrm{9}+{x}^{\mathrm{2}} }}−\frac{\mathrm{1}}{\mathrm{8}}=\mathrm{0} \\ $$$$\mathrm{8}{x}=\mathrm{6}\sqrt{\mathrm{9}+{x}^{\mathrm{2}} } \\ $$$$\mathrm{64}{x}^{\mathrm{2}} =\mathrm{36}\ast\mathrm{9}+\mathrm{36}{x}^{\mathrm{2}} \\ $$$$\mathrm{28}{x}^{\mathrm{2}} =\mathrm{36}\ast\mathrm{9} \\ $$$${x}=\frac{\mathrm{9}}{\:\sqrt{\mathrm{7}}} \\ $$$${T}\left(\mathrm{0}\right)=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${T}\left(\mathrm{8}\right)=\frac{\sqrt{\mathrm{73}}}{\mathrm{6}} \\ $$$${T}\left(\frac{\mathrm{9}}{\:\sqrt{\mathrm{7}}}\right)=\frac{\sqrt{\mathrm{9}+\frac{\mathrm{81}}{\mathrm{7}}}}{\mathrm{6}}+\mathrm{1}−\frac{\mathrm{9}}{\mathrm{8}\sqrt{\mathrm{7}}}=\frac{\sqrt{\frac{\mathrm{144}}{\mathrm{7}}}}{\mathrm{6}}+\mathrm{1}−\frac{\mathrm{9}}{\mathrm{8}\sqrt{\mathrm{7}}}=\sqrt{\mathrm{7}} \\ $$$${T}_{{min}} ={min}\left(\frac{\mathrm{3}}{\mathrm{2}},\frac{\sqrt{\mathrm{7}}}{\mathrm{8}}+\mathrm{1},\frac{\sqrt{\mathrm{73}}}{\mathrm{6}}\right)=\frac{\sqrt{\mathrm{7}}}{\mathrm{8}}+\mathrm{1} \\ $$$${so}\:{x}=\frac{\mathrm{9}}{\:\sqrt{\mathrm{7}}} \\ $$$$ \\ $$$$ \\ $$
Commented by liberty last updated on 10/Feb/21
$$\mathrm{yes}..= \\ $$