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maths

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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 - ,

    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 - ,

    sqrt[180*60] is not 80

  • maths - ,

    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 - ,

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

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