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#1 2007-05-02 07:07:56

dadon2
Guest

partial diff

Hey All,

how do determine the position and nature (i.e minimum, maximum or saddle point) of the stationary points of the function.

f(x,y)=x^2 - 4xy + y^3 + 4y

this is what I have done so far
dz/dx= 2x-4y
dz/dy=-4x+3y^2+4

d^2z/dxdy=-4

cheers

#2 2007-05-02 07:39:59

luca-deltodesco
Member
Registered: 2006-05-05
Posts: 1,470

Re: partial diff

and thats where i get stuck, i havn't covered this yet in my studies tongue


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#3 2007-05-02 07:56:24

luca-deltodesco
Member
Registered: 2006-05-05
Posts: 1,470

Re: partial diff

looked up eigenvalues, and this is how i work it out (but you don't have to trust me here, im new to this tongue)


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#4 2007-05-02 08:05:58

dadon2
Guest

Re: partial diff

thanks lol

don't worry about it. we are all learning here.

#5 2007-05-02 08:15:07

dadon2
Guest

Re: partial diff

how do you get ur x coordinates as 4/3 and 4

thanks

#6 2007-05-02 08:22:17

luca-deltodesco
Member
Registered: 2006-05-05
Posts: 1,470

Re: partial diff

stationary points are where both partial derivitaves are 0

produces simultaenous equation in x and y, note that ive rearranged partial in x, to give x = 2y, which ive substituted into partial in y to find values of y, and then plug back into the partial for x, to find value of x given the value of y


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#7 2007-05-02 08:57:44

dadon2
Guest

Re: partial diff

thanks

yes i can see what you have done.

but for

(4/3, 2/3), (4, 2)

where does the 4/3 and the 4 come from in the line above

do u substitute dem into a equation?

#8 2007-05-02 08:59:08

luca-deltodesco
Member
Registered: 2006-05-05
Posts: 1,470

Re: partial diff

i substitute the two values of yinto x = 2y, derived from the partial derivitave in x


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#9 2007-05-02 09:02:07

dadon2
Guest

Re: partial diff

oh I see!

thanks for that. I appreciate your help smile

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