Question Number 175623 by cortano1 last updated on 04/Sep/22
$$\:\mathrm{min}\:\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{2x}+\mathrm{5}}\:+\sqrt{\mathrm{4x}^{\mathrm{2}} −\mathrm{4x}+\mathrm{10}} \\ $$
Answered by greougoury555 last updated on 04/Sep/22
$$\:{let}\:\begin{cases}{{a}=\left(\mathrm{1}−{x}\right){i}+\mathrm{2}{j}}\\{{b}=\left(\mathrm{2}{x}−\mathrm{1}\right){i}+\mathrm{3}{j}}\end{cases} \\ $$$$\:\mid{a}\mid\:+\mid{b}\:\mid\:\geqslant\:\mid{a}\:+{b}\mid \\ $$$$\:{f}\left({x}\right)\:\geqslant\:\sqrt{{x}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} } \\ $$$$\:{holds}\:{when}\:\frac{\mathrm{1}−{x}}{\mathrm{2}{x}−\mathrm{1}}\:=\:\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\:\mathrm{4}{x}−\mathrm{2}\:=\:\mathrm{3}−\mathrm{3}{x}\:\Rightarrow{x}=\frac{\mathrm{5}}{\mathrm{7}} \\ $$$$\:{Min}\:{f}\left({x}\right)=\:\sqrt{\frac{\mathrm{25}}{\mathrm{49}}+\mathrm{25}}\:=\:\frac{\mathrm{25}\sqrt{\mathrm{2}}}{\mathrm{7}} \\ $$
Commented by infinityaction last updated on 04/Sep/22
$${what}\:{is}\:{i}\:{and}\:{j}\:?? \\ $$
Commented by mr W last updated on 04/Sep/22
$${i}\:{and}\:{j}\:{are}\:{unit}\:{vectors}\:{in}\:{the}\:{plane}. \\ $$
Commented by mr W last updated on 04/Sep/22
$${i}\:{think}\:{the}\:{solution}\:{is}\:{wrong}. \\ $$$$\:{from}\:{f}\left({x}\right)\:\geqslant\:\sqrt{{x}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} } \\ $$$${you}\:{can}\:{not}\:{get}\:{the}\:{minimum}\:{of}\:{f}\left({x}\right), \\ $$$${because}\:\sqrt{{x}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} }\:{is}\:{not}\:{a}\:{constant}, \\ $$$${but}\:{a}\:{function}\:{which}\:{in}\:{turn}\:{has}\:{its} \\ $$$${onw}\:{minimum}.\:{you}\:{can}\:{not}\:{say} \\ $$$${at}\:{x}=\frac{\mathrm{5}}{\mathrm{7}}\:\:\sqrt{{x}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} }\:{is}\:{minimum}. \\ $$
Commented by infinityaction last updated on 04/Sep/22
$${thanks}\:{sir} \\ $$