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

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

Answered by bobhans last updated on 12/Jul/20

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

error sir  (∂f/∂λ)=5x +2y−60 !

Commented bybemath last updated on 12/Jul/20

the equation 5x+12y−60=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))  .

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))

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))

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)))