Question Number 206987 by hardmath last updated on 02/May/24
$$\frac{\mathrm{120}}{\mid\mathrm{x}−\mathrm{1}\mid\:+\:\mid\mathrm{x}−\mathrm{3}\mid\:+\:\mid\mathrm{x}−\mathrm{5}\mid}\:\:\:\Rightarrow\:\:\:\mathrm{max}\:=\:? \\ $$
Answered by Frix last updated on 02/May/24
$$\frac{\mathrm{120}}{{f}\left({x}\right)}\:\mathrm{is}\:\mathrm{max}\:\Rightarrow\:{f}\left({x}\right)\:\mathrm{is}\:\mathrm{min} \\ $$$$\mathrm{4}\leqslant\mid{x}−\mathrm{1}\mid+\mid{x}−\mathrm{3}\mid+\mid{x}−\mathrm{5}\mid \\ $$$$\Rightarrow\:\mathrm{Answer}\:\mathrm{is}\:\frac{\mathrm{120}}{\mathrm{4}}=\mathrm{30} \\ $$
Commented by hardmath last updated on 02/May/24
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{dear}\:\mathrm{professor} \\ $$
Answered by A5T last updated on 03/May/24
$$\mid{x}−\mathrm{3}\mid=\mid\mathrm{3}−{x}\mid;\mid{x}−\mathrm{5}\mid=\mid\mathrm{5}−{x}\mid;\:\mid{a}\mid+\mid{b}\mid\geqslant\mid{a}+{b}\mid \\ $$$$\mid{x}−\mathrm{1}\mid+\mid\mathrm{3}−{x}\mid\geqslant\mid{x}−\mathrm{1}+\mathrm{3}−{x}\mid=\mathrm{2}…\left({i}\right) \\ $$$${Similarly},\:\mid{x}−\mathrm{1}\mid+\mid\mathrm{5}−{x}\mid\geqslant\mathrm{4}…\left({ii}\right) \\ $$$$\mid{x}−\mathrm{3}\mid+\mid\mathrm{5}−{x}\mid\geqslant\mathrm{2}…\left({iii}\right) \\ $$$$\left({i}\right)+\left({ii}\right)+\left({iii}\right)\Rightarrow\mid{x}−\mathrm{1}\mid+\mid{x}−\mathrm{3}\mid+\mid{x}−\mathrm{5}\mid\geqslant\mathrm{4} \\ $$$$\left({Equality}\:{when}\:{x}=\mathrm{3}\right) \\ $$$$\Rightarrow\frac{\mathrm{120}}{\mid{x}−\mathrm{1}\mid+\mid{x}−\mathrm{3}\mid+\mid{x}−\mathrm{5}\mid}\leqslant\frac{\mathrm{120}}{\mathrm{4}}=\mathrm{30}. \\ $$
Commented by hardmath last updated on 03/May/24
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{dear}\:\mathrm{professor} \\ $$