On critical points of a large function

Question

I am badly stuck at the following:$$f(x,y)=\frac{x^3}{3}-\frac{yx^2}{2}-\frac{\alpha x^2}{2}+\alpha yx+\frac{2y^3}{3}+y^2$$. The question is to find the value of $\alpha$ such that $f(x,y)$ has three distinct critical points simultaneously. One is given as $\alpha=0$, but how can we find the other, positive value of $\alpha$. Equating the first partial derivatives to zero leads to non-linear equations, which is tough to solve. How do we proceed? Thanks beforehand.

Answer 1

If you compute the partial derivatives you get $\frac{\partial f}{\partial x}(x,y)=(x-y)(x-\alpha)$ and $\frac{\partial f}{\partial y}(x,y)=-\frac12x^2+\alpha x+2y^2+2y$. So if you want critical points, from $\frac{\partial f}{\partial x}=0$ you have that either $x=y$ or $x=\alpha$. Plug these two values separately in the second equation.

Edit For $x=\alpha$ you get $\frac{\partial f}{\partial y}(\alpha,y)=\frac12\alpha^2+2y^2+2y=0$ for $y=\frac{1}{2}\sqrt{1-\alpha^{2}}-\frac{1}{2}$ and $y=-\frac{1}{2}% \sqrt{1-\alpha^{2}}-\frac{1}{2}$. Hence, if $|\alpha|<1$ the points $(\alpha,\frac{1}{2}\sqrt{1-\alpha^{2}}-\frac{1}{2})$ and $(\alpha,-\frac{1}{2}\sqrt{1-\alpha^{2}}-\frac{1}{2})$ are two critical points. When $x=y$ you get $\frac{\partial f}{\partial y}(x,x)=-\frac12x^2+\alpha x+2x^2+2x=x(\frac32x+\alpha+2)=0$ for $x=0$ and $x=-\frac23(\alpha+2)$ so you get $(0,0)$ and $(-\frac23(\alpha+2),-\frac23(\alpha+2))$.