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Question Number 81975 by TANMAY PANACEA last updated on 17/Feb/20

Commented by TANMAY PANACEA last updated on 17/Feb/20

$t h a n k y o u s i r$
Commented by mr W last updated on 17/Feb/20

$p l e a s e c h e c k y o u r m e t h o d s i r!$ $p o i n t (0, 9) i s w r o n g, s h o u l d b e (0, 7.5) .$ $p o i n t (6, 6) i s w r o n g, s h o u l d b e (5, 5) .$ $m i n i n u m 6 + 6 \sqrt{2} i s w r o n g, s h o u l d b e 6 \sqrt{5} .$
Commented by jagoll last updated on 17/Feb/20

Commented by mr W last updated on 17/Feb/20

$t o s a n t u s i r :$ $i t h i n k t h e e r r o r i s t h a t y o u a s s u m e d$ $t h a t t h e d i s t a n c e f r o m (3, 9) t o t h e$ $l i n e y = x m u s t b e m i n i m u m . b u t t h i s$ $i s n o t p r o v e d, a n d a l s o n o t t r u e .$
Commented by john santu last updated on 17/Feb/20

$h a h a, s o r r y t h e r e w a s w o r k . y e s$ $s i r, m y m e t h o d i s i n v a l i d$
	Answered by mr W last updated on 17/Feb/20

	Commented by mr W last updated on 17/Feb/20

	$P^{'} = m i r r o r i m a g e o f P a b o u t y - a x i s$ $A^{'} = m i r r o r i m a g e o f A a b o u t y = x$ $P^{″} = m i r r o r i m a g e o f P^{'} a b o u t y = x$ $p e r i m e t e r o f Δ P A B = P B + B A + A P$ $= P B + B A + A P^{'} = P B + {B A}^{'} + A^{'} P^{″}$ $m i n i m u m o f P B + {B A}^{'} + A^{'} P^{″} = P P^{″}$ $= g r e e n l i n e$ $= \sqrt{{(9 - 3)}^{2} + {(9 + 3)}^{2}} = 6 \sqrt{5} \approx 13.41$
	Commented by TANMAY PANACEA last updated on 17/Feb/20

	$e x c e l l e n t s i r . . .$
	Commented by mr W last updated on 17/Feb/20

	$e q n . o f P P^{″} :$ $\frac{y + 3}{9 + 3} = \frac{x - 9}{3 - 9}$ $\Rightarrow y = - 2 x + 15$ $i n t e r s e c t i o n w i t h l i n e y = x$ $y_{B} = - 2 x_{B} + 15 = x_{B}$ $\Rightarrow x_{B} = 5, y_{B} = 5 \Rightarrow p o i n t B (5, 5)$ $i n t e r s e c t i o n w i t h x - a x i s :$ $y_{A^{'}} = - 2 x_{A^{'}} + 15 = 0$ $\Rightarrow x_{A^{'}} = 7.5 \Rightarrow p o i n t A^{'} (7.5, 0)$ $\Rightarrow p o i n t A (0, 7.5)$
	Commented by mr W last updated on 17/Feb/20

	$i p r e f e r s u c h v i s i u a l s o l u t i o n . . .$
	Commented by john santu last updated on 17/Feb/20

	$p o i n t A (0, 7.5) w r o n g s i r .$ $t h e p o i n t i s n o t o n t h e l i n e y = x$ $. n o t a c c o r d a n c e w i t h p r o b l e m$
	Commented by mr W last updated on 17/Feb/20

	$c a l c u l u s m e t h o d :$ $s a y A (0, a), B (b, b)$ $p e r i m e t e r = p$ $p = \sqrt{3^{2} + {(9 - a)}^{2}} + \sqrt{{(3 - b)}^{2} + {(9 - b)}^{2}} + \sqrt{b^{2} + {(b - a)}^{2}}$ $p = \sqrt{90 - 18 a + a^{2}} + \sqrt{90 - 24 b + 2 b^{2}} + \sqrt{2 b^{2} - 2 a b + a^{2}}$ $\frac{\partial p}{\partial a} = \frac{- 9 + a}{\sqrt{90 - 18 a + a^{2}}} + \frac{- b + a}{\sqrt{2 b^{2} - 2 a b + a^{2}}} = 0$ $\Rightarrow \frac{9 - a}{\sqrt{90 - 18 a + a^{2}}} = \frac{a - b}{\sqrt{2 b^{2} - 2 a b + a^{2}}} . . . (i)$ $\frac{\partial p}{\partial b} = \frac{- 12 + 2 b}{\sqrt{90 - 24 b + 2 b^{2}}} + \frac{2 b - a}{\sqrt{2 b^{2} - 2 a b + a^{2}}} = 0$ $\frac{12 - 2 b}{\sqrt{90 - 24 b + 2 b^{2}}} = \frac{2 b - a}{\sqrt{2 b^{2} - 2 a b + a^{2}}} . . . (i i)$ $f r o m (i) a n d (i i) w e g e t a = \frac{15}{2}, b = 5 .$ $\Rightarrow A (0, \frac{15}{2}), B (5, 5)$
	Commented by mr W last updated on 17/Feb/20

	$s a n t u s i r :$ $p l e a s e c h e c k y o u r c o m m e n t a g a i n .$ $l o o k c a r e f u l l y a t t h e p i c t u r e :$ $o n t h e l i n e y = x i s t h e p o i n t B (5, 5) a n d$ $o n y - a x i s i s t h e p o i n t A (0, 7.5) .$ $w h a t i s w r o n g ?$
	Answered by ajfour last updated on 17/Feb/20
	$p=(√(9+(9−y)^2 ))+(√((h−3)^2 +(9−h)^2 )) +(√(h^2 +(y−h)^2 )) (∂p/∂y)=−((9−y)/(√(9+(9−y)^2 )))+((y−h)/(√(h^2 +(y−h)^2 )))=0 ⇒ (9−y)^2 h^2 =9(y−h)^2 ⇒ h(9−y)=3(y−h) 12h−hy−3y=0 ⇒ y=((12h)/(h+3)) ....(i) (∂p/∂h)=(((h−3)−(9−h))/(√((h−3)^2 +(9−h)^2 ))) +((h−(y−h))/(√(h^2 +(y−h)^2 ))) = 0 ⇒ ((2h−12)/(√((h−3)^2 +(9−h)^2 ))) =((2h−y)/(√(h^2 +(y−h)^2 ))) 4(h−6)^2 {h^2 +(y−h)^2 } =(2h−y)^2 {(h−3)^2 +(9−h)^2 } ⇒ 4(h−6)^2 {h^2 +(((12h)/(h+3))−h)^2 } = (2h−((12h)/(h+3)))^2 {2h^2 −24h+90} ⇒ 4(h−6)^2 {h^2 (h+3)^2 +h^2 (9−h)^2 } = 2×4h^2 (h−3)^2 (h^2 −12h+45) ⇒ (h−6)^2 (h^2 −6h+45) = (h−3)^2 (h^2 −12h+45) ⇒ 6h{2(h−3)^2 −(h−6)^2 } =(h^2 +45)(6h−27) ⇒ 2h(h^2 −18)=(h^2 +45)(2h−9) ⇒ −36h=−9h^2 +90h−9×45 ⇒ h^2 −14h+45=0 ⇒ (h−9)(h−5)=0 ⇒ h=5, 9 y=((12h)/(h+3))= ((15)/2), 9 so P(3,9) given , A(0,((15)/2)) on y-axis B (5,5) on y=x. minimum perimeter 6(√5) . Another set P(3,9) , A(0,9) , B(9,9) In this case perimeter is 18. Area is zero.$
	$p = \sqrt{9 + {(9 - y)}^{2}} + \sqrt{{(h - 3)}^{2} + {(9 - h)}^{2}}$ $+ \sqrt{h^{2} + {(y - h)}^{2}}$ $\frac{\partial p}{\partial y} = - \frac{9 - y}{\sqrt{9 + {(9 - y)}^{2}}} + \frac{y - h}{\sqrt{h^{2} + {(y - h)}^{2}}} = 0$ $\Rightarrow {(9 - y)}^{2} h^{2} = 9 {(y - h)}^{2}$ $\Rightarrow h (9 - y) = 3 (y - h)$ $12 h - h y - 3 y = 0$ $\Rightarrow y = \frac{12 h}{h + 3} . . . . (i)$ $\frac{\partial p}{\partial h} = \frac{(h - 3) - (9 - h)}{\sqrt{{(h - 3)}^{2} + {(9 - h)}^{2}}}$ $+ \frac{h - (y - h)}{\sqrt{h^{2} + {(y - h)}^{2}}} = 0$ $\Rightarrow \frac{2 h - 12}{\sqrt{{(h - 3)}^{2} + {(9 - h)}^{2}}}$ $= \frac{2 h - y}{\sqrt{h^{2} + {(y - h)}^{2}}}$ $4 {(h - 6)}^{2} {h^{2} + {(y - h)}^{2}}$ $= {(2 h - y)}^{2} {{(h - 3)}^{2} + {(9 - h)}^{2}}$ $\Rightarrow 4 {(h - 6)}^{2} {h^{2} + {(\frac{12 h}{h + 3} - h)}^{2}}$ $= {(2 h - \frac{12 h}{h + 3})}^{2} {2 h^{2} - 24 h + 90}$ $\Rightarrow 4 {(h - 6)}^{2} {h^{2} {(h + 3)}^{2} + h^{2} {(9 - h)}^{2}}$ $= 2 \times 4 h^{2} {(h - 3)}^{2} (h^{2} - 12 h + 45)$ $\Rightarrow {(h - 6)}^{2} (h^{2} - 6 h + 45)$ $= {(h - 3)}^{2} (h^{2} - 12 h + 45)$ $\Rightarrow 6 h {2 {(h - 3)}^{2} - {(h - 6)}^{2}}$ $= (h^{2} + 45) (6 h - 27)$ $\Rightarrow 2 h (h^{2} - 18) = (h^{2} + 45) (2 h - 9)$ $\Rightarrow - 36 h = - 9 h^{2} + 90 h - 9 \times 45$ $\Rightarrow h^{2} - 14 h + 45 = 0$ $\Rightarrow (h - 9) (h - 5) = 0$ $\Rightarrow h = 5, 9$ $y = \frac{12 h}{h + 3} = \frac{15}{2}, 9$ $s o$ $P (3, 9) g i v e n, A (0, \frac{15}{2}) o n y - a x i s$ $B (5, 5) o n y = x .$ $m i n i m u m p e r i m e t e r 6 \sqrt{5} .$ $A n o t h e r s e t$ $P (3, 9), A (0, 9), B (9, 9)$ $I n t h i s c a s e p e r i m e t e r i s 18 .$ $A r e a i s z e r o .$
	Commented by mr W last updated on 17/Feb/20

	$t h a n k s f o r c o n f i r m i n g s i r!$
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