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Multivariable Calculus: Max/min values inside closed interval?

  1. Dec 12, 2013 #1
    Hi!

    My question is:
    Find maximum and minimum values of the function:

    f(x,y) = 2x-y+x^2+y^2

    when x^2+y^2 ≤ 4

    I would like to solve this without using Lagrange method.

    I get
    x=-1 and y=1/2 when using partial derivative and set it equql to 0.

    I can see that the maximum value must be at the edge of the circle, therefore x^2+y^2 = 4

    f(x,y) = 2x-y+x^2+y^2 =
    2x-y+4 =
    2x - sqrt( 4 - x^2 ) + 4

    I derivative and get:

    2 + x/(sqrt(4-x^2))=0

    x=-4/sqrt(5)

    Hmm, is this really the right method?
     
  2. jcsd
  3. Dec 12, 2013 #2

    LCKurtz

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    You mean ##2x \color{red}{\pm} \sqrt{4-x^2}+4##. You get two ##y## values for each ##x## on the circle.

    I didn't check your last steps but, yes, that method should work. Don't let the square roots scare you.
     
  4. Dec 12, 2013 #3

    vela

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    So (-1, 1/2) is one candidate.

    Your work is fine, but when you solve for y in terms of x, you get two roots ##y=\pm\sqrt{4-x^2}##. You need to consider the second solution still.
     
  5. Dec 12, 2013 #4
    Thanks both of you!

    I forgotten the second root.
    Now I can continue and see what I get for answer.
     
  6. Dec 12, 2013 #5
    It wasn't as easy as I thought.

    I get x=4/(sqrt(5) and x=-4/sqrt(5)

    It seems that x=4/(sqrt(5) is correct when I studie the the 3D picture http://www.wolframalpha.com/input/?i=f%28x%2Cy%29%3D2x-y%2Bx^2%2By^2

    Do I need to studie the egde of the circle using 2csot, 2sint? I increase the value by increase x.
     
  7. Dec 12, 2013 #6
    Ok let's see if I'm on the right track.

    I solve for y:

    x^2+y^2=4

    y=±sqrt(4-y^2)

    I put it in the equation and differentiate:

    f(y)=2*±sqrt(4-y^2)-y+4=
    f´(y)=±(2 y)/sqrt(4-y^2)-1

    When I set it equal to 0:

    y= -(2/sqrt(5))
    y= (2/sqrt(5))


    If I sum it up I have three candidates:

    Point I (-1, 1/2)
    Point II (4/sqrt(5)), (-2/sqrt(5))
    Point III (-4/sqrt(5)), (2/sqrt(5))


    If I insert each point into the original equation I get:

    Point I = -5/4 =-1.25
    Point II = 2(2+sqrt(5)) ≈ 8.47
    Point III = 4-2*sqrt(5) ≈-0.47

    Point I = minimum
    Point II = maximum

    Is this solution good enough?
     
  8. Dec 12, 2013 #7

    vela

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    Looks good.
     
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