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Boyle22
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This is a calc 3 problem dealing with partial derivatives in 3space:
I know that for a function to have a max at a specified point (a,b):
Fx(a,b)=0
Fy(a,b)=0
D>0 where D=Fxx(a,b)Fyy(a,b)-[Fxy(a,b)]^2
D can not equal 0
Fxx(a,b)<0
For a min at specified point:
Fx(a,b)=0
Fy(a,b)=0
D>0 where D=Fxx(a,b)Fyy(a,b)-[Fxy(a,b)]^2
D can not equal 0
Fxx(a,b)>0So I've been spending the past few hours trying to figure out this problem and I'm not exactly sure how to go about doing it. I've just been making up functions by trial and error and haven't been getting anywhere really.
The one function that comes closest that I came up with was:
(x^2)(y^3)-(x^4)(y^6)+1/2(x^4)(y^6)
since x and y partials are both equal to 0 at point (1,1) and Fxx partial is negative... but when i plug in for D I get 0.
So what is the proper way of approaching this problem?
Find a function (continuous, smooth) which has a local max at (1,1) and a local minimum at (-1,-1) and prove.
I know that for a function to have a max at a specified point (a,b):
Fx(a,b)=0
Fy(a,b)=0
D>0 where D=Fxx(a,b)Fyy(a,b)-[Fxy(a,b)]^2
D can not equal 0
Fxx(a,b)<0
For a min at specified point:
Fx(a,b)=0
Fy(a,b)=0
D>0 where D=Fxx(a,b)Fyy(a,b)-[Fxy(a,b)]^2
D can not equal 0
Fxx(a,b)>0So I've been spending the past few hours trying to figure out this problem and I'm not exactly sure how to go about doing it. I've just been making up functions by trial and error and haven't been getting anywhere really.
The one function that comes closest that I came up with was:
(x^2)(y^3)-(x^4)(y^6)+1/2(x^4)(y^6)
since x and y partials are both equal to 0 at point (1,1) and Fxx partial is negative... but when i plug in for D I get 0.
So what is the proper way of approaching this problem?
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