Posted by **John** on Tuesday, March 12, 2013 at 4:48am.

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

- 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[(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 :)

- 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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