maths

posted by .

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

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Math(Roots)

    sqrt(24) *I don't really get this stuff.Can somebody please help me?
  2. math,algebra,help

    Directions are simplify by combining like terms. x radiacal 18 -3 radical 8x^2 can someone show me how to do these types of problems. thanks I cant determine the second term. For the first, I think you meant x sqrt(18) which reduces …
  3. Math

    How do you find a square root of a number that's not a perfect square?
  4. math,correction

    simplify: sqrt ((7)/(100)) my work: 7 = 1sqrt (7) 100= 10 sqrt (1) (1)/(10) sqrt (7) my answer: (sqrt (7))/(10) Your answer is right but your statements: 7 = 1sqrt (7) 100= 10 sqrt (1) make no sense. how can sqrt(7)=7 ?
  5. Mathematics

    sqrt 6 * sqrt 8 also sqrt 7 * sqrt 5 6.92820323 and 5.916079783 So you can see the steps — sqrt 6 * sqrt 8 = sqrt 48 sqrt 7 * sqrt 5 = sqrt 35 I hope this helps a little more. Thanks for asking.
  6. Math Help please!!

    Could someone show me how to solve these problems step by step.... I am confused on how to fully break this down to simpliest terms sqrt 3 * sqrt 15= sqrt 6 * sqrt 8 = sqrt 20 * sqrt 5 = since both terms are sqrt , you can combine …
  7. Calculus

    Please look at my work below: Solve the initial-value problem. y'' + 4y' + 6y = 0 , y(0) = 2 , y'(0) = 4 r^2+4r+6=0, r=(16 +/- Sqrt(4^2-4(1)(6)))/2(1) r=(16 +/- Sqrt(-8)) r=8 +/- Sqrt(2)*i, alpha=8, Beta=Sqrt(2) y(0)=2, e^(8*0)*(c1*cos(0)+c2*sin(0))=c2=2 …
  8. Math/Calculus

    Solve the initial-value problem. Am I using the wrong value for beta here, 2sqrt(2) or am I making a mistake somewhere else?
  9. Algebra

    Find the largest possible real value of \sqrt{(x-20)(y-x)} + \sqrt{(140-y)(20-x)} + \sqrt{(x-y)(y-140)} subject to -40\leq x \leq 100 and -20\leq y \leq 200.
  10. Algebra

    Evaluate sqrt7x (sqrt x-7 sqrt7) Show your work. sqrt(7)*sqrt(x)-sqrt(7)*7*sqrt(7) sqrt(7*x)-7*sqrt(7*7) sqrt(7x)-7*sqrt(7^2) x*sqrt 7x-49*x ^^^ would this be my final answer?

More Similar Questions