Menu Close

Question-145759




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

Leave a Reply

Your email address will not be published. Required fields are marked *