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Absolute Extrema: 2 variables

  1. Mar 9, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the absolute extrema of the function on the set D.

    f(x,y)= x^2 + 4y^2 + 3x -1

    D= {(x,y) l x^2 + y^2 ≤ 4}

    2. Relevant equations



    3. The attempt at a solution

    The only thing I've done so far was find the critical point. I found f-sub x = 2x+3 and f-suby= 8y giving me a critical point of (-3/2,0), correct?

    Using that critical point, I found f(-3/2,0) = -1

    After that, I don't know how to proceed. I know -2 ≤ x ≤ 2 and -2 ≤ y ≤ 2 but how do I find the other critical numbers with that info? I'm unsure because the closed boundary graph is a circle and not a square, box, or triangular region.
     
  2. jcsd
  3. Mar 10, 2013 #2

    ehild

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    f(-3/2) is not equal to -1.

    The absolute extreme can be on the perimeter. Find the critical point with the condition that x2+y2=4.

    ehild
     
  4. Mar 10, 2013 #3
    Ah, you're right, thank you. For that, I got -13/4.




    So set y=√(4-x^2) ? and find f(x,√(4-x^2)). Then find values of x and use the result to find y. Then find critical numbers? Because that's what I did but it didn't seem right.
     
    Last edited: Mar 10, 2013
  5. Mar 10, 2013 #4

    Ray Vickson

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    You have found the global minimum of f, subject to g = x^2 + y^2 <= 4. (although your f is wrong while your (x,y) is correct). However, an extremum can also be a maximum, and you have not yet found that.
     
  6. Mar 10, 2013 #5

    ehild

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    That is right.

    Show your work.

    ehild
     
  7. Mar 10, 2013 #6
    Thanks, guys! I got it now. I forgot to take the derivative of f(x,√(4-x^2)).

    Instead of finding the derivative, I found x and y of f(x,√(4-x^2)) and plugged it into the original function which was wrong.
     
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