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Question Number 150056 by ajfour last updated on 09/Aug/21

Commented by ajfour last updated on 09/Aug/21

$I f t h e b l u e r e g i o n i s a s q u a r e$ $o f m a x i m u m a r e a, f i n d a$ $a n d s i d e s o f t h e s q u a r e .$
	Answered by mr W last updated on 09/Aug/21

	$y = {(x - a)}^{2} = x$ $x^{2} - (2 a + 1) x + a^{2} = 0$ $s = x = \frac{2 a + 1 - \sqrt{4 a + 1}}{2}$ $\frac{d s}{d a} = 0$ $a - \frac{2}{\sqrt{4 a + 1}} = 0$ $a^{2} = \frac{4}{4 a + 1}$ $\frac{1}{a^{3}} - \frac{1}{4 a} - 1 = 0$ $\frac{1}{a} = \sqrt[3]{\frac{\sqrt{1293}}{72} + \frac{1}{2}} - \sqrt[3]{\frac{\sqrt{1297}}{72} - \frac{1}{2}}$ $\Rightarrow a \approx 0.9232$
	Commented by ajfour last updated on 09/Aug/21

	$t h a n k u s i r, e a s y 4 u .$
	Commented by ajfour last updated on 09/Aug/21

	$n o i t i s n o t s o, s i r .$ $f o r a g i v e n a, t h e a r e a m i g h t$ $b e m a x . f o r a r e c t a n g l e;$ $A n d f o r t h i s m a x i m u m a r e a$ $r e c t a n g l e, t o b e a s q u a r e a s w e l l$ $, t h e r e h a s t o b e s o m e s p e c i f i c$ $a v a l u e, a n d h e n c e s v a l u e .$
	Commented by mr W last updated on 09/Aug/21

	$b u t i n f a c t t h e r e i s n o a b s o l u t e$ $m a x i m u m o r m i n i m u m . s i n c e$ $s \to \infty w h e n a \to \infty$ $s \to 0 w h e n a \to 0$
	Commented by mr W last updated on 09/Aug/21

	$t h e u p p e r r i g h t c o r n e r o f t h e s q u a r e$ $o n t h e p a r a b o l a i s t h e i n t e r s e c t i o n$ $p o i n t o f y = {(x - a)}^{2} a n d y = x .$ $w e s e e t h e i n t e r s e c t i o n p o i n t a l w a y s$ $e x i s t s f o r a n y v a l u e o f a f r o m 0 t i l l \infty .$
	Commented by mr W last updated on 09/Aug/21

	$i f y o u m e a n t h e c a s e t h a t t h e s q u a r e$ $h a p p e n s t o b e t h e m a x i m u m r e c t a n g l e,$ $t h e n t h e r e i s s u c h a v a l u e f o r a .$
	Answered by mr W last updated on 09/Aug/21

	$s a y t h e u p p e r r i g h t c o r n e r o f t h e$ $r e c t a n g l e i s (p, {(p - a)}^{2}) .$ $a r e a o f r e c t a n g l e i s$ $A = p {(p - a)}^{2}$ $\frac{d A}{d p} = {(p - a)}^{2} + 2 p (p - a) = 0$ $p = \frac{a}{3}$ $s u c h t h a t t h e m a x i m u m r e c t a n g l e$ $i s a s q u a r e,$ $p = {(p - a)}^{2}$ $\frac{a}{3} = {(\frac{a}{3} - a)}^{2}$ $\Rightarrow a = \frac{3}{4}$
	Commented by ajfour last updated on 09/Aug/21

	$Y e s, t h i s i s a n s w e r t o m y$ $q u e s t i o n, s i r . t h a n k s f o r$ $a l l e f f o r t s . r e a l l y g o o d .$
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