a^2 + b^2 = 1/4
where a and b are integral values what would be the value of a and b for ab to be minimum?
A) 1/sqrt8
B) -1/sqrt8
C) -1/8
D) 0

• 2ab =(a+b)^2 - (a^2+b^2)
  => ab = 1/2*(a+b)^2 - 1/2*(a^2+b^2)
  => ab = 1/2*(a+b)^2 - 1/8
  
  as, 1/2*(a+b)^2 > 0 for any a,b
  for ab to be minm 1/2*(a+b)^2 = 0 => a+b=0 => b=-a
  & minm value of ab = -1/8
  a^2 + b^2 = 1/4
  => 2*a^2 = 1/4
  => a^2 = 1/8
  => a = 1/sqrt8 , -1/sqrt8
  
  however, a^2 + b^2 = 1/4 is not possible for any integral values of a & b.
  
  minm value of ab = -1/8 for (a,b)= (1/sqrt8 , -1/sqrt8) or (-1/sqrt8 , 1/sqrt8)
  
• 10 years agoHelpfull: Yes(16) No(2)
• a^2+b^2=1/4
  we can write above as
  => (a+b)^2-2ab=1/4
  =>2ab=(a+b)^2-1/4
  =>ab= [(a+b)^2-1/4]/2
  now,
  ab is minimum when a+b=0 =>b= -a
  so ab= -1/8
  => a(-a)=-1/8
  =>-a^2=-1/8
  =>a^2=1/8
  =>a= +- 1/sqrt8
  but in this problem a= 1/sqrt8
  because in question it was mention that b
• 10 years agoHelpfull: Yes(9) No(0)
• c) -1/8
  here ab=1/2*(a+b)^2-1/8
  so the 1st term i.e. 1/2*(a+b)^2 can never be negative so minimum value of this should be 0 for this. so min value for ab is -1/8
  But the fact is a and b cannot be integers bcoz then a^2+b^2 can never be eaqual to 1/4
• 10 years agoHelpfull: Yes(1) No(0)
• A)1/sqrt(8)
• 10 years agoHelpfull: Yes(1) No(0)
• 1/sqrt8 is the answer
• 10 years agoHelpfull: Yes(1) No(0)
• C is the correct answer
• 10 years agoHelpfull: Yes(0) No(0)
• for this que if we go through the options
  then, for A>
  b^2 = 1/4 - 1/8
  b = 1/ sqrt8 so, ab = 1/8
  for B>
  b^2 = 1/4- 1/8
  b = 1/sqrt8 so, ab = -1/8
  for C>
  b^2 = 1/4 - 1/64
  b = sqrt(15/64) = sqrt(15)/8
  so, ab = -sqrt(15)/64
  for D>
  ab = 0
  so from all the options
  A > D > B > C
  so for ab to be minimum ans = C> -1/8
• 10 years agoHelpfull: Yes(0) No(0)
• now continue from my above answer
  
  ant it was mention in question that b
• 10 years agoHelpfull: Yes(0) No(0)
• for two nos x and y ..A.M(x,y)>= G.M.(x,y)
  put x= a^2,y =b ^2..
  (a^2+b^2)/2 >= sqrt(a^2*b^2);
  1/8 >=sqrt(a^2*b^2);
  sqrt(1/8)>=|ab|;
  hence ab belongs to {-sqrt(-1/8),sqrt(1/8)};
  
  
• 10 years agoHelpfull: Yes(0) No(0)
• (a+b)^2=a^2+b^2+2ab
  (a+b)^2>=0
  a^2+b^2+2ab>=0
  1/4+2ab>=0
  ab>= - 1/8
• 9 years agoHelpfull: Yes(0) No(0)