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anonymous

  • one year ago

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    @ganeshie8

  3. Loser66
    • one year ago
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    I change it to x, y instead of x1, x2 for convenience. \(f(x,y) = 3xy-x^3-y^3\), it is a symmetric function, hence \(f_x = 3y-3x^2\\f_{xx}= -6x\\f_y= 3x-3y^2\\f_{yy}= -6y\\f_{xy}=f_{yx}=3\)

  4. Loser66
    • one year ago
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    The rule to figure out whether it is max or min: \(if~~f_{xx} <0, f_{yy}<0 ~~at~~(a,b)\rightarrow\) it is max \(if~~f_{xx}>0, f_{yy}>0~~at~~(a,b)\rightarrow\) it is min

  5. anonymous
    • one year ago
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    Do I disregard the f_xy and f_yx?

  6. Loser66
    • one year ago
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    We will use it them if the conditions above doesn't fit.

  7. anonymous
    • one year ago
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    and both f_xx and f_yy would be equal to zero only when x and y are equal to 0

  8. anonymous
    • one year ago
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    which would make f_x and f_y critical points, but I dont know if thyre a max or min because the second order derivative at those points isnt negative or positive

  9. Loser66
    • one year ago
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    Oh, I am sorry, to find max/min, we use \(f_x=0,f_y=0\) not \(f_{xx}, f_{yy}\)

  10. Loser66
    • one year ago
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    Tell me , what do you get for the point?

  11. anonymous
    • one year ago
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    so for the points that make f_y equal to zero, did you get x=0 or 1^(1/3)?

  12. Loser66
    • one year ago
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    how? \(f_x= 3y-3x^2=3(y-x^2)=0\) iff \(y-x^2=0\) \(f_y = 3x -3y^2= 3(x-y^2)=0\) iff \(x -y^2=0\) Hence we have \(y=x^2\\x=y^2\) only (0,0) and (1,1) are solutions for both them, right?

  13. anonymous
    • one year ago
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    oh i added something wrong. I see now. Thanks! but only (1,1) would be a solution correct? because it says in the problem that x1 and x2 (x and Y) are greater than .5

  14. Loser66
    • one year ago
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    I don't know yet. I have to check one by one. Now (0,0) consider \(f_{xx}f_{yy}-f^2_{xy}\) at (0,0) , what do you have?

  15. Loser66
    • one year ago
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    oh, ok, I forgot that condition. YOu are correct

  16. Loser66
    • one year ago
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    now, check the same expression for (1,1)

  17. anonymous
    • one year ago
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    wait whats the f^2_xy?

  18. Loser66
    • one year ago
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    \(f_{xy}=3\rightarrow f_{xy}^2=9\)

  19. anonymous
    • one year ago
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    is it just the squared value of f_xy? I apologize for all the questions but the first class was yesterday so im coming into this knowing nothing

  20. anonymous
    • one year ago
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    ok thought so

  21. anonymous
    • one year ago
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    What is the purpose of FxxFyy-F^2xy?

  22. anonymous
    • one year ago
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    I got 27 btw

  23. Loser66
    • one year ago
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    Yes, so it is >0, right? --> (1,1) either max or min. Now, consider if it is max or min by :

  24. Loser66
    • one year ago
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    \(f_{xx}\) at (1,1)= ? \(f_{yy}\) at (1,1) =?

  25. anonymous
    • one year ago
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    They're both negative meaning its a max

  26. Loser66
    • one year ago
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    yyyyyyyyyyyyyyyyyyyyyyyyyyyes

  27. Loser66
    • one year ago
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    You got it.

  28. anonymous
    • one year ago
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    Wait what was the FxxFyy-Fxy thing I havent seen that yet

  29. anonymous
    • one year ago
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    Was that just to determine the value of the point?

  30. anonymous
    • one year ago
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    Where did the formula come from?

  31. Loser66
    • one year ago
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    we have fxxfyy-f^2xy to find out the saddle point or max/min point if it is <0, the point is saddle point if it is >0 the point is either max or min.

  32. anonymous
    • one year ago
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    Whats a Saddle point? again, sorry for all the questions

  33. anonymous
    • one year ago
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    Nevermind i got it. Thanks so much

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