maths
posted by John .
Find the largest possible value of
sqrt[(x20)(yx)] + sqrt[(140y)(20x)] + sqrt[(xy)(y140)] if
40<=x<=100 and 20<=y<=200

Lets assume that x>y
then x>20 (as x>y minimum value of y is 20)
see the first term
sqrt[(x20)(yx)]
as x>20 and x>y
the term inside the sqrt becomes negative.
Hence x!>y (not greater than y)
similarly you can prove that y is not greater than x.
Hence x=y
put x=y in the expression
only middle term is left.
sqrt[(140y)(20x)]
to maximise we put x=40
sqrt[(140(40))(20(40))]
=80
hence the answer is 80 :) 
sqrt[180*60] is not 80

We have the limitation that 20 <= y <= 200, so we can't say that x = y = 40. Instead, we have to use x = y = 20 to maximize it, so sqrt(160 * 40) is 80.

Oh yeah!!! that's was a typo :P