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Question Number 54061 by behi83417@gmail.com last updated on 28/Jan/19

Commented by mr W last updated on 29/Jan/19

$C (t, \frac{1}{t})$ $D (\frac{1}{t}, t)$ $C D = \sqrt{2} (t - \frac{1}{t}) = 2$ $t^{2} - \sqrt{2} t - 1 = 0$ $\Rightarrow t = \frac{1 + \sqrt{3}}{\sqrt{2}}$ $C B = \sqrt{2} (\frac{1 + \sqrt{3}}{\sqrt{2}} + \frac{\sqrt{2}}{1 + \sqrt{3}}) = 2 \sqrt{3}$ $A r e a o f r e c t a n g l e = 2 \times 2 \sqrt{3} = 4 \sqrt{3}$ $t h e r e i s o n l y o n e p o s s i b l e r e c t a n g l e .$ $b u t q u e s t i o n a s k s f o r m a x i m u m r e c t a n g l e .$
Commented by behi83417@gmail.com last updated on 28/Jan/19

$A B C D, i s a r e c t a n g l e a n d i t s d i a g o n a l s$ $m e e t e a c h o t h e r a t c e n t e r p o i n t o f :$ $x y - p l a n e .$ $i f : A B = 2 (c o n s t a n t), f i n d m a x i m u m$ $a r e a o f r e c t a n g l e .$
Commented by behi83417@gmail.com last updated on 28/Jan/19

$?$
Commented by behi83417@gmail.com last updated on 28/Jan/19

$s i r A j f o u r! t h a n k y o u f o r a t t e n t i o n .$ $b u t t h i s i s n o t c l e a r f o r m e .$ $s o m e e x p l a i n t i o n p l e a s e ?$ $A B, i s t h e w i d t h o f r e c t a n g l e a n d$ $l o c a t e d o n : y = \frac{1}{x} .$
Commented by behi83417@gmail.com last updated on 29/Jan/19

$t h a n k s i n a d v a n c e d e a r m a s t e r .$ $w h a t a b o u t a n s w e r i f : A B C D b e a$ $p a r a l l e l o g r a m ?$
Commented by behi83417@gmail.com last updated on 29/Jan/19

$I {d o n}^{'} t r e c i v e a n y n o t i f i c a t i o n s f r o m$ $t i n k u t a r a a p p . a m i a l o n e ?$
Commented by mr W last updated on 29/Jan/19

$i {d o n}^{'} t g e t n o t i f i c a t i o n s e i t h e r . s o i$ ${d o n}^{'} t k n o w i f s o m e o n e h a s c o m m e n t e d$ $m y p o s t s .$
Commented by peter frank last updated on 30/Jan/19

$t r u e t h e s a m e t o m e i d o n t g e t n o t i f i c a t i o n .$
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