## anonymous one year ago Image attached

1. anonymous

2. anonymous

@ganeshie8

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

4. 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

5. anonymous

Do I disregard the f_xy and f_yx?

6. Loser66

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

7. anonymous

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

8. 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

9. Loser66

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

10. Loser66

Tell me , what do you get for the point?

11. anonymous

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

12. 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?

13. 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

14. 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?

15. Loser66

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

16. Loser66

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

17. anonymous

wait whats the f^2_xy?

18. Loser66

$$f_{xy}=3\rightarrow f_{xy}^2=9$$

19. 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

20. anonymous

ok thought so

21. anonymous

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

22. anonymous

I got 27 btw

23. Loser66

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

24. Loser66

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

25. anonymous

They're both negative meaning its a max

26. Loser66

yyyyyyyyyyyyyyyyyyyyyyyyyyyes

27. Loser66

You got it.

28. anonymous

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

29. anonymous

Was that just to determine the value of the point?

30. anonymous

Where did the formula come from?

31. 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.

32. anonymous

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

33. anonymous

Nevermind i got it. Thanks so much