I looked up some random textbook on internet for multivariable functions. A=d2f/(dx)^2=2 B=d2f/(dxdy)=-2 C=d2f/(dy)^2=6y We need to calculate the discriminant B^2-AC I) x=y=0 then B°2-AC=4>0 not a point of extrema II) x=y=2/3 B^2-AC=4-(2)(6*2/3)=-4 If A0 sonwe got our local minimum point. Because there exists y such that f(2/3,y)
I found some books about multivariable surface calculus but my memory of doing the subject is not great I know you usually want df/dx=0 and df/dy=0 plus some additional conditions to know if it is Mon/max or saddle point In this case df/dx=0 2x-2y=0 so X=y And df/Dy=0 -2x+3y^2=0 So A is true in a sense it is necessary but not sufficient condition B is false C not sure as (df/dx^2)^2=2 and (df/dy^2)^2=3y =0 at (0,0) D obviously false E f(X,y)=(x-y)^2-y^2+y^3=(x-y)^2+y^2(y-1) (x-y)^2 has a valley at X=y and y^2(y-1) is negative when y
I was just thinking that f(X,y) = g(X,y)+h(y) where h(y)=y^3-y^2 then h'(y)=0 means 3y^2-2y=0 or y=0 y=2/3 where y=0 is a max for cubic and y=2/3 is a minimum In this context (2/3,2/3) makes sense to be minimum for the whole function I think A is false because X=y also contains the absolute extrema. This problem was written in such a way you have to mond the grammar
I looked up some random textbook on internet for multivariable functions.
A=d2f/(dx)^2=2
B=d2f/(dxdy)=-2
C=d2f/(dy)^2=6y
We need to calculate the discriminant B^2-AC
I) x=y=0 then B°2-AC=4>0 not a point of extrema
II) x=y=2/3 B^2-AC=4-(2)(6*2/3)=-4
If A0 sonwe got our local minimum point.
Because there exists y such that f(2/3,y)
I found some books about multivariable surface calculus but my memory of doing the subject is not great
I know you usually want df/dx=0 and df/dy=0 plus some additional conditions to know if it is Mon/max or saddle point
In this case df/dx=0 2x-2y=0 so X=y
And df/Dy=0 -2x+3y^2=0
So A is true in a sense it is necessary but not sufficient condition
B is false
C not sure as (df/dx^2)^2=2 and (df/dy^2)^2=3y =0 at (0,0)
D obviously false
E f(X,y)=(x-y)^2-y^2+y^3=(x-y)^2+y^2(y-1)
(x-y)^2 has a valley at X=y and y^2(y-1) is negative when y
I was just thinking that f(X,y) = g(X,y)+h(y) where h(y)=y^3-y^2 then h'(y)=0 means 3y^2-2y=0 or y=0 y=2/3 where y=0 is a max for cubic and y=2/3 is a minimum
In this context (2/3,2/3) makes sense to be minimum for the whole function
I think A is false because X=y also contains the absolute extrema. This problem was written in such a way you have to mond the grammar
I’m sure that they consider an absolute extreme to also be a relative extreme. The intended answer in the test is option A.