Posted by John on Tuesday, March 12, 2013 at 4:48am.
Find the largest possible value of
sqrt[(x20)(yx)] + sqrt[(140y)(20x)] + sqrt[(xy)(y140)] if
40<=x<=100 and 20<=y<=200

maths  Saurabh D, Wednesday, March 13, 2013 at 1:23am
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 :)

maths  Anonymous, Wednesday, March 13, 2013 at 3:14am
sqrt[180*60] is not 80

maths  Anonymous, Thursday, March 14, 2013 at 10:53pm
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.

maths  Saurabh D, Sunday, March 17, 2013 at 9:02am
Oh yeah!!! that's was a typo :P
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