find the minimum value
Calculus
Level
2
if A+B+C+D=a and A-B+C-D=b then what is the minimum value of A^2+B^2+C^2+D^2 ?
(a^2+b^2)/4
(a^2+b^2)/2
(a^2-b^2)/4
ab/4
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1 solution
A+B+C+D=a A-B+C-D=b now A+B=a-B-D,,put in eq 2 we get (a-b)/2=B+D. then we get A+C=(a+b)/2 squaring both the eq...we get B^2+D^2+2BD=(a-b)^2/4 A^2+C^2+2CA=(a+b)^2/4 adding both and neglecting 2BD and 2CA as asked minimum value...we get (a^2+b^2)/4