Question

Image attached

Answer 1

@ganeshie8

Answer 2

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
\(f_x = 3y-3x^2\\f_{xx}= -6x\\f_y= 3x-3y^2\\f_{yy}= -6y\\f_{xy}=f_{yx}=3\)

Answer 3

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

Answer 4

Do I disregard the f_xy and f_yx?

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

Tell me , what do you get for the point?

Answer 10

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

Answer 11

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?

Answer 12

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

Answer 13

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?

Answer 14

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

Answer 15

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

Answer 16

wait whats the f^2_xy?

Answer 17

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

Answer 18

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

Answer 19

ok thought so

Answer 20

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

Answer 21

I got 27 btw

Answer 22

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

Answer 23

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

Answer 24

They're both negative meaning its a max

Answer 25

yyyyyyyyyyyyyyyyyyyyyyyyyyyes

Answer 26

You got it.

Answer 27

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

Answer 28

Was that just to determine the value of the point?

Answer 29

Where did the formula come from?

Answer 30

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.

Answer 31

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

Answer 32

Nevermind i got it. Thanks so much