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

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

$${given}\:\mathrm{5}{x}+\mathrm{12}{y}\:=\:\mathrm{60} \\ $$$${min}\:{value}\:{of}\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} } \\ $$

Answered by bobhans last updated on 12/Jul/20

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

error sir (∂f/∂λ)=5x +2y−60 !

$$\mathrm{error}\:\mathrm{sir}\:\:\frac{\partial\mathrm{f}}{\partial\lambda}=\mathrm{5x}\:+\mathrm{2y}−\mathrm{60}\:! \\ $$

Commented by bemath last updated on 12/Jul/20

the equation 5x+12y−60=0 sir it typo

$${the}\:{equation}\:\mathrm{5}{x}+\mathrm{12}{y}−\mathrm{60}=\mathrm{0}\:{sir} \\ $$$${it}\:{typo}\: \\ $$$$ \\ $$$$ \\ $$

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)) .

$${let}\:{x}={r}\mathrm{cos}\:\theta\:,\:{y}={r}\mathrm{sin}\:\theta \\ $$$$\Rightarrow\:\:\:{r}=\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} } \\ $$$$\:\:\:\mathrm{13}\left(\frac{\mathrm{5}{x}}{\mathrm{13}}+\frac{\mathrm{12}{y}}{\mathrm{13}}\right)=\mathrm{60} \\ $$$$\left\{{if}\:\:\:\mathrm{sin}\:\alpha=\frac{\mathrm{5}}{\mathrm{13}}\:,\:\:\mathrm{cos}\:\alpha=\frac{\mathrm{12}}{\mathrm{13}}\right. \\ $$$$\left.\Rightarrow\:\:\:\:\:\mathrm{tan}\:\alpha=\frac{\mathrm{5}}{\mathrm{12}}\:\right\} \\ $$$$\:\:\mathrm{13}{r}\left(\mathrm{sin}\:\alpha\mathrm{cos}\:\theta+\mathrm{cos}\:\alpha\mathrm{sin}\:\theta\right)=\mathrm{60} \\ $$$$\Rightarrow\:\:\mathrm{13}{r}\mathrm{sin}\:\left(\alpha+\theta\right)=\mathrm{60} \\ $$$$\Rightarrow\:\:\:\:{r}=\frac{\mathrm{60}}{\mathrm{13sin}\left(\alpha+\theta\right)} \\ $$$${and}\:\:{as}\:\:{r}>\mathrm{0}\:\:\:\:\Rightarrow\:\:{r}_{{min}} =\frac{\mathrm{60}}{\mathrm{13}} \\ $$$${when}\:\:\theta=\mathrm{2}{n}\pi+\frac{\pi}{\mathrm{2}}−\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{5}}{\mathrm{12}}\:\:. \\ $$

Answered by maths mind last updated on 12/Jul/20

M(x,y)∈(D): 5x+12y−60=0 (√(x^2 +y^2 ))=d(O,M) min d(O,M)=d(O,M′) M′ ptojection of O over (d) d(0,M′)=((∣0.5+12.0−60∣)/(√(5^2 +12^2 )))=((60)/(13))

$${M}\left({x},{y}\right)\in\left({D}\right):\:\mathrm{5}{x}+\mathrm{12}{y}−\mathrm{60}=\mathrm{0} \\ $$$$\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }={d}\left({O},{M}\right) \\ $$$${min}\:{d}\left({O},{M}\right)={d}\left({O},{M}'\right) \\ $$$${M}'\:{ptojection}\:{of}\:{O}\:{over}\:\left({d}\right) \\ $$$${d}\left(\mathrm{0},{M}'\right)=\frac{\mid\mathrm{0}.\mathrm{5}+\mathrm{12}.\mathrm{0}−\mathrm{60}\mid}{\sqrt{\mathrm{5}^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} }}=\frac{\mathrm{60}}{\mathrm{13}} \\ $$

Answered by mr W last updated on 12/Jul/20

say (√(x^2 +y^2 ))=D x^2 +y^2 =D^2 x^2 +(5−((5x)/(12)))^2 =D^2 ((169)/(144))x^2 −((50)/(12))x+25−D^2 =0 Δ=(((50)/(12)))^2 −4×((169)/(144))(25−D^2 )=0 D^2 =((3600)/(169)) ⇒D=((60)/(13))

$${say}\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }={D} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={D}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +\left(\mathrm{5}−\frac{\mathrm{5}{x}}{\mathrm{12}}\right)^{\mathrm{2}} ={D}^{\mathrm{2}} \\ $$$$\frac{\mathrm{169}}{\mathrm{144}}{x}^{\mathrm{2}} −\frac{\mathrm{50}}{\mathrm{12}}{x}+\mathrm{25}−{D}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Delta=\left(\frac{\mathrm{50}}{\mathrm{12}}\right)^{\mathrm{2}} −\mathrm{4}×\frac{\mathrm{169}}{\mathrm{144}}\left(\mathrm{25}−{D}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${D}^{\mathrm{2}} =\frac{\mathrm{3600}}{\mathrm{169}} \\ $$$$\Rightarrow{D}=\frac{\mathrm{60}}{\mathrm{13}} \\ $$

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)))

$$\mathrm{Apply}\:\mathrm{Cauchy}−\mathrm{Shward}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{60}^{\mathrm{2}} =\left(\mathrm{5x}+\mathrm{12y}\right)^{\mathrm{2}} \leqslant\left(\mathrm{5}^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} \right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \geqslant\frac{\mathrm{60}^{\mathrm{2}} }{\mathrm{5}^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} }=\frac{\mathrm{60}^{\mathrm{2}} }{\mathrm{13}^{\mathrm{2}} }\Rightarrow\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }\geqslant\frac{\mathrm{60}}{\mathrm{13}} \\ $$$$\mathrm{the}\:\mathrm{equality}\:\mathrm{ocurrs}\:\mathrm{if}\:\mathrm{and}\:\mathrm{only}\:\mathrm{if} \\ $$$$\begin{cases}{\frac{\mathrm{x}}{\mathrm{5}}=\frac{\mathrm{y}}{\mathrm{12}}}\\{\mathrm{5x}+\mathrm{12y}=\mathrm{60}}\end{cases}\Leftrightarrow\begin{cases}{\mathrm{x}=\frac{\mathrm{300}}{\mathrm{169}}}\\{\mathrm{y}=\frac{\mathrm{720}}{\mathrm{169}}}\end{cases} \\ $$$$\boldsymbol{\mathrm{Thus}},\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{y}}^{\mathrm{2}} }\:\boldsymbol{\mathrm{has}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{smallest}}\:\boldsymbol{\mathrm{value}} \\ $$$$\boldsymbol{\mathrm{equal}}\:\frac{\mathrm{60}}{\mathrm{13}}\:\boldsymbol{\mathrm{when}}\:\left(\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{y}}\right)=\left(\frac{\mathrm{300}}{\mathrm{169}};\frac{\mathrm{720}}{\mathrm{169}}\right) \\ $$

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