What can be the maximum and minimum values of (A+B) and (A-B) ?

Asked by  | 18th Jun, 2013, 11:39: AM

Expert Answer:

A+B = sqrt(A^2 +B^2 + 2ABcos(theta))
A-B = sqrt(A^2 +B^2 - 2ABcos(theta))
Maximum value of cos(theta) can be 1 and minimum value can be -1. 
 
hence for maximum value of cos(theta) 
A+B = sqrt(A^2 +B^2 + 2AB) = A+B
A-B = sqrt(A^2 +B^2 - 2AB) = A-B
For minimum value of cos(theta)
A+B = sqrt(A^2 +B^2 -2AB) = A-B
A+B = sqrt(A^2 +B^2 - 2AB(-1)) = A+B
 
So, the maximum values of both A+B and A-B will be A+B (but for different values of cos(theta))
and the minimum values for both A+B and A-B would be A-B (again for different values of cos(theta))

Answered by  | 19th Jun, 2013, 07:35: AM

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