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- anonymous

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- anonymous

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- anonymous

@ganeshie8

- 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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## More answers

- 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

Do I disregard the f_xy and f_yx?

- Loser66

We will use it them if the conditions above doesn't fit.

- anonymous

and both f_xx and f_yy would be equal to zero only when x and y are equal to 0

- 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

Oh, I am sorry, to find max/min, we use \(f_x=0,f_y=0\) not \(f_{xx}, f_{yy}\)

- Loser66

Tell me , what do you get for the point?

- anonymous

so for the points that make f_y equal to zero, did you get x=0 or 1^(1/3)?

- 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

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

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

oh, ok, I forgot that condition. YOu are correct

- Loser66

now, check the same expression for (1,1)

- anonymous

wait whats the f^2_xy?

- Loser66

\(f_{xy}=3\rightarrow f_{xy}^2=9\)

- 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

ok thought so

- anonymous

What is the purpose of FxxFyy-F^2xy?

- anonymous

I got 27 btw

- Loser66

Yes, so it is >0, right? --> (1,1) either max or min.
Now, consider if it is max or min by :

- Loser66

\(f_{xx}\) at (1,1)= ?
\(f_{yy}\) at (1,1) =?

- anonymous

They're both negative meaning its a max

- Loser66

yyyyyyyyyyyyyyyyyyyyyyyyyyyes

- Loser66

You got it.

- anonymous

Wait what was the FxxFyy-Fxy thing
I havent seen that yet

- anonymous

Was that just to determine the value of the point?

- anonymous

Where did the formula come from?

- 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

Whats a Saddle point? again, sorry for all the questions

- anonymous

Nevermind i got it. Thanks so much

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