If 5x+12y=60 then find the minimum value of sqrt(x^2+y^2)?

• put x=0.we get y=5
  so,sqrt(0^2+5^2)=5
  and
  if y=0 than x=12
  thn sqrt(12^2+0^2)=12
  so the minimum value is 5 is the ans.
• 11 years agoHelpfull: Yes(35) No(9)
• 5x+12y=60
  0r (5/13)x + (12/13)y = 60/13
  let x=rcosA, y=rsinA so that r=sqrt(x^2 +y^2)
  or r(cosA*cosB+sinA.sinB)=60/13 (if 5/13=cosB then sinB=12/13)
  or r*cos(A-B)=60/13
  or r=(60/13)/cos(A-B)
  FOR r=sqrt(x^2 +y^2) to be minm cos(A-B)maxm=1
  so minm value of sqrt(x^2+y^2)=60/13
  
• 11 years agoHelpfull: Yes(10) No(6)
• ANSWER is 4.60
  Put x=1 then y=4.5 now value of (x^2+y^2) is (1^2+4.5^2)=(1+20.25)=21.25
  now Sqrt(21.25)=4.60 which is minimum among all the values.
  So ANSWER is 4.60
• 11 years agoHelpfull: Yes(9) No(4)
• 5*0+12*5=60
  ans:5
• 11 years agoHelpfull: Yes(4) No(3)
• (5x+12y)^2+(12x−5y)^2=(13^2)*(x^2+y^2).
  Since 5x+12y is given, we minimize x^2+y^2 by minimizing (12x−5y)^2. The minimum value of this is 0. It follows that the minimum value of 60/13.
• 11 years agoHelpfull: Yes(3) No(3)
• 5x+12y=60
  y=5 ans 25
  
• 11 years agoHelpfull: Yes(1) No(4)
• x=0,y=5
  ans is5
• 11 years agoHelpfull: Yes(0) No(0)