Examine $f(x,y)=4x^2-xy+4y^2+x^3y+xy^3-4$ for extreme values My attempt: $$f_x=8x-y+3x^2y+y^3=0$$ $$f_y=-x+8y+x^3+3xy^2=0$$ Adding $f_x$ and $f_y$,I got $7x+7y+(x+y)^3=0$ $$(x+y)[7+(x+y)^2]=0$$ How do I get the stationary points from this?
The solution is given as $(0,0),(3/2,-3/2),(-3/2,3/2)$
On
The equation you derived came from $f_x+f_y=0$. On the other hand, notice that
\begin{align} f_x-f_y &=\left[(-x)^3+3(-x)^2y+3(-x)y^2+y^3\right]+9x-9y\\ &=(y-x)^3-9(y-x) \end{align}
Let $u=x+y$ and $v=y-x$. It follows that if $(x,y)$ is a stationary point the following system is satisfied:
$$\left\{ \begin{array}{l} u^3+7u=0\\ v^3-9v=0 \end{array} \right.$$
The first equation yields $u(u^2+7)=0$, so we must have $u=0$ since $u$ must be real.
The second yields $v(v^2-9)=0$, so we must have $v=0$ or $v=\pm3$.
Now it's a matter of recovering the pairs $(x,y)$ from the pairs $(u,v)$, and checking whether or not they are indeed stationary points.
$\qquad (1)$ $u=v=0:$ Corresponds to $x=y=0$, which is easily checked to satisfy $\nabla f=0$.
$\qquad (2)$ $u=0, v=3:$ Corresponds to $y=3/2$ and $x=-3/2$.
$\qquad (3)$ $u=0, v=-3:$ Corresponds to $y=-3/2$ and $x=3/2$.
The product of two real numbers is zero if and only if at least one of them is zero. Since $7+(x+y)^2>0$, the only solutions are given by $x+y=0$. Hence you must insert the condition $x+y=0$ into your equations $\partial f/\partial x=0$ and $\partial f/\partial y=0$.