I have the following function $f:\mathbb{R}^{2}\rightarrow \mathbb{R}$ defined by
$$f(x,y)=xy(x+y-1).$$
Using the property that the gradient in the stationary points is zero I calculated the four stationary points:
$$(0,0), (0,1), (1,0), (1/3,1/3)$$
in my homework assignment I now have to prove that the function has only these stationary points no more and no less. I don't really know how to do this can anyone help?
$$f_x(x,y) = 2xy + y^2 - y=y(2x+y-1)$$
$$f_y(x,y) = x^2 + 2xy+ x=x(x+2y+1)$$
Hence the stationary point must satisfy
($y=0$ or $2x+y-1=0$) and ($x=0$ or $x+2y+1=0$)
Consider each of the $4$ combinations, we are always have a linear system of which the system is non-singular (as the constraints are not parallel to each other), hence the solution for each system is unique. Hence there are at most $4$ solutions.
Since you have found $4$ solutions, there are exactly $4$ solutions.
from the first equation $$f_x=2xy+y^2-y$$ we get $$y(2x+y-1)=0$$ from here we get $$y=0$$ or $$y=1-2x$$ this can be plug into the second equation and this a poynomial $$5\,{x}^{2}-5\,x+1=0$$ and this has at least two Solutions.
