Finding stationary points of a function when substitution method fails

Question

I have a function, $f(x,y)=x^2+2xy+y^2+\alpha x+\beta y$ where $\beta$ and $\alpha$ are parameters.

I'm asked to find the stationary points of the function so naturally, I tried to find where the derivative might be 0.

I solved the partial derivatives to be: ($2x+2y+\alpha$,$2x+2y+\beta$) respectively.

This gives me:

1. $2x+2y+\alpha=0$
2. $2x+2y+\beta=0$

Now, the assumption that I'm allowed is that $\alpha=\beta$ which means I rewrite the two equations as (for example):

1. $2x+2y+\alpha=0$
2. $2x+2y+\alpha=0$

However the problem I'm having is solving the system now to find the stationary points.

The substitution method does not work, nor will the addition method. I've also tried to convert it into row echelon format:

\begin{matrix} 1 \ 1 \ \frac{1}{2} \ 0 \\ 0 \ 0 \ 0 \ 0 \ \end{matrix}

but this doesn't really produce any solution as well.

Can you guys provide maybe some other methods to find the stationary points or possibly tell me if I've done something incorrectly?

Thanks in advance.

Answer 1

You haven't failed. Your stationary points are the intersection of the line: $y = \dfrac{-\alpha - 2x}{2}$ with your surface $z = f(x,y)$.

Here is a plot in 3D for $\alpha = \beta = 1$:

The plane is defined by the line $y = \dfrac{-\alpha - 2}{2}$ and the surface is $z = x^2 + 2x y + y^2 + x + y$.