given 5x+12y = 60 min value of (√(x^2 +y^2 ))


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Question Number 103040 by bobhans last updated on 12/Jul/20

$g i v e n 5 x + 12 y = 60$ $m i n v a l u e o f \sqrt{x^{2} + y^{2}}$
	Answered by bobhans last updated on 12/Jul/20

	Commented by mathmax by abdo last updated on 12/Jul/20

	$error sir \frac{\partial f}{\partial λ} = 5 x + 2 y - 60!$
	Commented by bemath last updated on 12/Jul/20

	$t h e e q u a t i o n 5 x + 12 y - 60 = 0 s i r$ $i t t y p o$
	Answered by ajfour last updated on 12/Jul/20
	$let x=rcos θ , y=rsin θ ⇒ r=(√(x^2 +y^2 )) 13(((5x)/(13))+((12y)/(13)))=60 {if sin α=(5/(13)) , cos α=((12)/(13)) ⇒ tan α=(5/(12)) } 13r(sin αcos θ+cos αsin θ)=60 ⇒ 13rsin (α+θ)=60 ⇒ r=((60)/(13sin(α+θ))) and as r>0 ⇒ r_(min) =((60)/(13)) when θ=2nπ+(π/2)−tan^(−1) (5/(12)) .$
	$l e t x = r \cos θ, y = r \sin θ$ $\Rightarrow r = \sqrt{x^{2} + y^{2}}$ $13 (\frac{5 x}{13} + \frac{12 y}{13}) = 60$ ${i f \sin α = \frac{5}{13}, \cos α = \frac{12}{13}$ $\Rightarrow \tan α = \frac{5}{12}}$ $13 r (\sin α \cos θ + \cos α \sin θ) = 60$ $\Rightarrow 13 r \sin (α + θ) = 60$ $\Rightarrow r = \frac{60}{13 \sin (α + θ)}$ $a n d a s r > 0 \Rightarrow r_{m i n} = \frac{60}{13}$ $w h e n θ = 2 n π + \frac{π}{2} - \tan^{- 1} \frac{5}{12} .$
	Answered by maths mind last updated on 12/Jul/20

	$M (x, y) \in (D) : 5 x + 12 y - 60 = 0$ $\sqrt{x^{2} + y^{2}} = d (O, M)$ $m i n d (O, M) = d (O, M^{'})$ $M^{'} p t o j e c t i o n o f O o v e r (d)$ $d (0, M^{'}) = \frac{∣ 0.5 + 12.0 - 60 ∣}{\sqrt{5^{2} + 12^{2}}} = \frac{60}{13}$
	Answered by mr W last updated on 12/Jul/20

	$s a y \sqrt{x^{2} + y^{2}} = D$ $x^{2} + y^{2} = D^{2}$ $x^{2} + {(5 - \frac{5 x}{12})}^{2} = D^{2}$ $\frac{169}{144} x^{2} - \frac{50}{12} x + 25 - D^{2} = 0$ $Δ = {(\frac{50}{12})}^{2} - 4 \times \frac{169}{144} (25 - D^{2}) = 0$ $D^{2} = \frac{3600}{169}$ $\Rightarrow D = \frac{60}{13}$
	Answered by 1549442205 last updated on 13/Jul/20
	$Apply Cauchy−Shward we have 60^2 =(5x+12y)^2 ≤(5^2 +12^2 )(x^2 +y^2 ) ⇒x^2 +y^2 ≥((60^2 )/(5^2 +12^2 ))=((60^2 )/(13^2 ))⇒(√(x^2 +y^2 ))≥((60)/(13)) the equality ocurrs if and only if { (((x/5)=(y/(12)))),((5x+12y=60)) :}⇔ { ((x=((300)/(169)))),((y=((720)/(169)))) :} Thus,(√(x^2 +y^2 )) has the smallest value equal ((60)/(13)) when (x,y)=(((300)/(169));((720)/(169)))$
	$Apply Cauchy - Shward we have$ $60^{2} = {(5 x + 12 y)}^{2} ⩽ (5^{2} + 12^{2}) (x^{2} + y^{2})$ $\Rightarrow x^{2} + y^{2} ⩾ \frac{60^{2}}{5^{2} + 12^{2}} = \frac{60^{2}}{13^{2}} \Rightarrow \sqrt{x^{2} + y^{2}} ⩾ \frac{60}{13}$ $the equality ocurrs if and only if$ ${\begin{cases} \frac{x}{5} = \frac{y}{12} \\ 5 x + 12 y = 60 \end{cases} \Leftrightarrow {\begin{cases} x = \frac{300}{169} \\ y = \frac{720}{169} \end{cases}$ $Thus, \sqrt{x^{2} + y^{2}} has the smallest value$ $equal \frac{60}{13} when (x, y) = (\frac{300}{169}; \frac{720}{169})$
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