Find the largest possible area of trapezoid that can be drawn under the x−axis so that one of its bases is on the x−axis and the other two vertices are on the curve y=x^2 −9


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Question Number 185453 by cortano1 last updated on 22/Jan/23

$F i n d t h e l a r g e s t p o s s i b l e a r e a$ $o f t r a p e z o i d t h a t c a n b e d r a w n$ $u n d e r t h e x - a x i s s o t h a t o n e$ $o f i t s b a s e s i s o n t h e x - a x i s$ $a n d t h e o t h e r t w o v e r t i c e s a r e$ $o n t h e c u r v e y = x^{2} - 9$
Commented by mr W last updated on 22/Jan/23

$q u e s t i o n i s w r o n g . w i t h g i v e n$ $c o n d i t i o n s t r a p e z o i d c a n b e \infty l a r g e .$
Commented by cortano1 last updated on 22/Jan/23

$t h e t r a p e z o i d u n d e r x - a x i s$
Commented by cortano1 last updated on 22/Jan/23

Commented by mathlove last updated on 22/Jan/23

$a n y a p p u s e d f o r t h e g r a p h ?$
Commented by mr W last updated on 22/Jan/23

$b u t y o u r q u e s t i o n d e s c r i b e s a$ $t r a p e z o i d l i k e t h i s :$
Commented by mr W last updated on 22/Jan/23

Commented by mr W last updated on 22/Jan/23

$i w r o n g o r y o u w r o n g ?$
Commented by cortano1 last updated on 22/Jan/23

$i t h i n k t h e q u e s t i o n r e q u i r e d$ $b a s e o f t r a p e z o i d m u s t$ $b e o n c u r v e y = x^{2} - 9$
Commented by mr W last updated on 22/Jan/23

${t h a t}^{'} s e x a c t l y w h a t i s h o w e d a b o v e :$ $o n e b a s e i s o n t h e x - a x i s, t h e$ $o t h e r b a s e i s o n t h e c u r v e .$ $b a s e s a r e t h e t w o p a r a l l e l s i d e s!$
	Answered by mahdipoor last updated on 22/Jan/23
	$get −3≤b≤a≤3 S=((∣a^2 −9∣×∣b^2 −9∣×∣a−b∣)/2)=(((9−a^2 )(9−b^2 )(a−b))/2) { (((∂S/∂b)=((9−a^2 )/2)(−9+3b^2 −2ba)=0)),(((∂S/∂a)=((9−b^2 )/2)(9−3a^2 +2ab)=0)) :} ⇒ { ((b= 3 or −(√(9/5)))),((a=−3 or (√(9/5)) )) :} ⇒ S=0 S=((72)/(25))(√(9/5)) ⇒⇒S(max)=((72)/(25))(√(9/5))=((216)/(25(√5)))$
	$g e t - 3 ⩽ b ⩽ a ⩽ 3$ $S = \frac{∣ a^{2} - 9 ∣ \times ∣ b^{2} - 9 ∣ \times ∣ a - b ∣}{2} = \frac{(9 - a^{2}) (9 - b^{2}) (a - b)}{2}$ ${\begin{cases} \frac{\partial S}{\partial b} = \frac{9 - a^{2}}{2} (- 9 + 3 b^{2} - 2 b a) = 0 \\ \frac{\partial S}{\partial a} = \frac{9 - b^{2}}{2} (9 - 3 a^{2} + 2 a b) = 0 \end{cases}$ $\Rightarrow {\begin{cases} b = 3 o r - \sqrt{\frac{9}{5}} \\ a = - 3 o r \sqrt{\frac{9}{5}} \end{cases}$ $\Rightarrow S = 0 S = \frac{72}{25} \sqrt{\frac{9}{5}}$ $\Rightarrow\Rightarrow S (m a x) = \frac{72}{25} \sqrt{\frac{9}{5}} = \frac{216}{25 \sqrt{5}}$
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