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
sqrt[(x-20)(y-x)] + sqrt[(140-y)(20-x)] + sqrt[(x-y)(y-140)] if
-40<=x<=100 and -20<=y<=200
Answers
Saurabh D
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 :)
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 :)
Anonymous
sqrt[180*60] is not 80
Anonymous
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.
Saurabh D
Oh yeah!!! that's was a typo :P