anonymous
  • anonymous
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Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
anonymous
  • anonymous
@ganeshie8
Loser66
  • Loser66
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 \(f_x = 3y-3x^2\\f_{xx}= -6x\\f_y= 3x-3y^2\\f_{yy}= -6y\\f_{xy}=f_{yx}=3\)

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Loser66
  • Loser66
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
anonymous
  • anonymous
Do I disregard the f_xy and f_yx?
Loser66
  • Loser66
We will use it them if the conditions above doesn't fit.
anonymous
  • anonymous
and both f_xx and f_yy would be equal to zero only when x and y are equal to 0
anonymous
  • anonymous
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
Loser66
  • Loser66
Oh, I am sorry, to find max/min, we use \(f_x=0,f_y=0\) not \(f_{xx}, f_{yy}\)
Loser66
  • Loser66
Tell me , what do you get for the point?
anonymous
  • anonymous
so for the points that make f_y equal to zero, did you get x=0 or 1^(1/3)?
Loser66
  • Loser66
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?
anonymous
  • anonymous
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
Loser66
  • Loser66
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?
Loser66
  • Loser66
oh, ok, I forgot that condition. YOu are correct
Loser66
  • Loser66
now, check the same expression for (1,1)
anonymous
  • anonymous
wait whats the f^2_xy?
Loser66
  • Loser66
\(f_{xy}=3\rightarrow f_{xy}^2=9\)
anonymous
  • anonymous
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
anonymous
  • anonymous
ok thought so
anonymous
  • anonymous
What is the purpose of FxxFyy-F^2xy?
anonymous
  • anonymous
I got 27 btw
Loser66
  • Loser66
Yes, so it is >0, right? --> (1,1) either max or min. Now, consider if it is max or min by :
Loser66
  • Loser66
\(f_{xx}\) at (1,1)= ? \(f_{yy}\) at (1,1) =?
anonymous
  • anonymous
They're both negative meaning its a max
Loser66
  • Loser66
yyyyyyyyyyyyyyyyyyyyyyyyyyyes
Loser66
  • Loser66
You got it.
anonymous
  • anonymous
Wait what was the FxxFyy-Fxy thing I havent seen that yet
anonymous
  • anonymous
Was that just to determine the value of the point?
anonymous
  • anonymous
Where did the formula come from?
Loser66
  • Loser66
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.
anonymous
  • anonymous
Whats a Saddle point? again, sorry for all the questions
anonymous
  • anonymous
Nevermind i got it. Thanks so much

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