MATEMATİK I - Deneme Sınavı - 1

Firstly, for critical points the first derivatives df(x,y)/dx=f_x and df(x,y)/dy=f_y must be equal to zero simultaneously. Namely f_x=f_y=0.

Since f(x,y)=x²y²-x²-y²?

f_x=2xy²-2x=0 so 2x(y²-1)=0 which means either x=0 or y=1 or y=-1

f_y=2yx²-2y=0 so 2y(x²-1)=0 which means either y=0 or x=1 or x=-1

But since both equations must be satisfied simultaneously we have 5 critical points:

(0,0), (-1,1), (1,-1), (1,1) and (-1,-1).

And also for a local maximum the following conditions must be satisfied simultaneously

I. f_xx<0

II.0xx*f_yy-f_xy*f_yx 

Thus we now have to find the second partial derivatives for an evaluation:

f_xx=2y²-2

f_yy=2x²-2

f_xy=f_yx =4yx

Among the critical points for only point (0,0) the first condition is satisfied (namely f_xx<0). Lets check whether this point also satisfies the second condition:

f_xx*f_yy-f_xy*f_yx =(2y²-2)*(2x²-2)-16x²y²=(-2)*(-2)-0=4

since 0<4 the second condition is also satisfied. Therfore point (0,0) is a local maximum.