find minimum value of function y=(√((x+6)^2 +25)) +(√((x−6)^2 +121))


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Question Number 114922 by bemath last updated on 22/Sep/20

$f i n d m i n i m u m v a l u e o f f u n c t i o n$ $y = \sqrt{{(x + 6)}^{2} + 25} + \sqrt{{(x - 6)}^{2} + 121}$
	Answered by john santu last updated on 22/Sep/20

	$y o u w a n t t o f i n d t h e p o i n t o n t h e$ $x - a x i s s u c h t h a t t h e s u m o f i t s$ $d i s t a n c e f r o m t h e p o i n t s (- 6, 5) a n d$ $(6, 11) i s m i n i m a l .$ $c o n s i d e r t h e s y m e t r i c p o i n t o f (- 6, 5)$ $w i t h r e s p e c t t o t h e a x i s, t h a t i s (- 6, - 5) .$ $t h e l i n e t h r o u g h (- 6, - 5) a n d (6, 11)$ $h a s e q u a t i o n 4 x - 3 y + 9 = 0$ $a n d i t i n t e r s e c t s t h e x - a x i s a t$ $x = - \frac{9}{4} . T h e m i n i m u m v a l u e$ $i s t h e r e f o r e y = \sqrt{{(- \frac{9}{4} + 6)}^{2} + 25} + \sqrt{{(- \frac{9}{6} - 6)}^{2} + 121}$ $y = \frac{25}{4} + \frac{55}{4} = 20 ∴$
	Commented by bemath last updated on 22/Sep/20

	$g a v e k u d o s$
	Answered by bobhans last updated on 22/Sep/20
	$with calculus y = (√((x+6)^2 +25)) + (√((x−6)^2 +121)) y ′=(((x+6))/( (√((x+6)^2 +25)))) + (((x−6))/( (√((x−6)^2 +121)))) = 0 (x+6)(√((x−6)^2 +121)) +(x−6)(√((x+6)^2 +25)) = 0 (x+6)(√((x−6)^2 +121)) = (6−x)(√((x+6)^2 +25)) ((√((x−6)^2 +121))/( (√((x+6)^2 +25)))) = ((6−x)/(x+6)) (((x−6)^2 +121)/((x+6)^2 +25)) = (((x−6)^2 )/((x+6)^2 )) (x^2 −36)^2 +121(x+6)^2 =(x^2 −36)^2 +25(x−6)^2 (11x+66)^2 =(5x−30)^2 (16x+36)(6x+96)=0 { ((x=−(9/4))),((x=−16)) :} for x=−(9/4) y=(√((−(9/4)+6)^2 +25)) +(√((−(9/4)−6)^2 +121)) y=(√((625)/(16))) + (√((3025)/(16))) = ((25)/4)+((55)/4)=20 for x=−16 y=(√((−16+6)^2 +25)) +(√((−16−6)^2 +121)) y=(√(125)) +(√(605)) = 35.78 therefore minimum value is 20$
	$w i t h c a l c u l u s$ $y = \sqrt{{(x + 6)}^{2} + 25} + \sqrt{{(x - 6)}^{2} + 121}$ $y^{'} = \frac{(x + 6)}{\sqrt{{(x + 6)}^{2} + 25}} + \frac{(x - 6)}{\sqrt{{(x - 6)}^{2} + 121}} = 0$ $(x + 6) \sqrt{{(x - 6)}^{2} + 121} + (x - 6) \sqrt{{(x + 6)}^{2} + 25} = 0$ $(x + 6) \sqrt{{(x - 6)}^{2} + 121} = (6 - x) \sqrt{{(x + 6)}^{2} + 25}$ $\frac{\sqrt{{(x - 6)}^{2} + 121}}{\sqrt{{(x + 6)}^{2} + 25}} = \frac{6 - x}{x + 6}$ $\frac{{(x - 6)}^{2} + 121}{{(x + 6)}^{2} + 25} = \frac{{(x - 6)}^{2}}{{(x + 6)}^{2}}$ ${(x^{2} - 36)}^{2} + 121 {(x + 6)}^{2} = {(x^{2} - 36)}^{2} + 25 {(x - 6)}^{2}$ ${(11 x + 66)}^{2} = {(5 x - 30)}^{2}$ $(16 x + 36) (6 x + 96) = 0$ ${\begin{cases} x = - \frac{9}{4} \\ x = - 16 \end{cases}$ $f o r x = - \frac{9}{4}$ $y = \sqrt{{(- \frac{9}{4} + 6)}^{2} + 25} + \sqrt{{(- \frac{9}{4} - 6)}^{2} + 121}$ $y = \sqrt{\frac{625}{16}} + \sqrt{\frac{3025}{16}} = \frac{25}{4} + \frac{55}{4} = 20$ $f o r x = - 16$ $y = \sqrt{{(- 16 + 6)}^{2} + 25} + \sqrt{{(- 16 - 6)}^{2} + 121}$ $y = \sqrt{125} + \sqrt{605} = 35.78$ $t h e r e f o r e m i n i m u m v a l u e i s 20$
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