partial derivative and stationary points
Consider f(x,y) = 2x^2y+xy+2/3(y^3)
Calculated partial derivates fx=4xy+u , fy=2(x^2)+x+2(y^2)
fxx=4y , fxy=4x+1 , fyx=4x+1 , fyy =4y
used partial df/dx =0 = 4xy+y solve y(4x+1)=0 => y=0 x=-1/4
(Headbang)how do I solve partial df/dy= 0 = 2(x^2)+x+2(y^2).
The answer for all stationary points is
(0,0) saddle point; (-1/2, 0 ) saddle point , ( -1/4, 1/4) local min. : (-1/4, -1/4) local min.
Stuck any advice????? (Surprised)
Re: partial derivative and stationary points
substituting the values of x and y you can solve df/dy