Question

Find the largest possible value of
sqrt[(x-20)(y-x)] + sqrt[(140-y)(20-x)] + sqrt[(x-y)(y-140)] if
-40<=x<=100 and -20<=y<=200

Answer 1

Lets assume that x>y
then x>-20 (as x>y minimum value of y is -20)
see the first term
sqrt[(x-20)(y-x)]
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[(140-y)(20-x)]
to maximise we put x=-40
sqrt[(140-(-40))(20-(-40))]
=80
hence the answer is 80 :)

Answer 2

sqrt[180*60] is not 80

Answer 3

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.

Answer 4

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