Question Number 145759 by mathdanisur last updated on 07/Jul/21
Commented by mr W last updated on 07/Jul/21
$$\sqrt{{ab}}\leqslant\frac{{a}+{b}}{\mathrm{2}}={x} \\ $$$$\Rightarrow{ab}\leqslant{x}^{\mathrm{2}} ={maximum} \\ $$
Commented by mathdanisur last updated on 07/Jul/21
$${Thankyou}\:{Ser},\:{a}\neq{b}.? \\ $$
Answered by mr W last updated on 07/Jul/21
$${say}\:{a}={x}+{k} \\ $$$${then}\:{b}={x}−{k} \\ $$$${ab}=\left({x}+{k}\right)\left({x}−{k}\right)={x}^{\mathrm{2}} −{k}^{\mathrm{2}} \\ $$$$\left({ab}\right)_{{max}} ={x}^{\mathrm{2}} \:{when}\:{k}=\mathrm{0},\:{i}.{e}.\:{a}={b}. \\ $$$${if}\:{a}\neq{b},\:{then}\:{k}\geqslant\mathrm{1},\:\left({ab}\right)_{{max}} ={x}^{\mathrm{2}} −\mathrm{1} \\ $$
Commented by mathdanisur last updated on 07/Jul/21
$${cool}\:{Ser},\:{thankyou} \\ $$