Optimal points of $f(x,y)=x^2 + y^2 + \beta xy + x + 2y$ I am self-learning basic optimization theory and algorithms from &quot;An Introduction to Optimization&quot; by Chong and Zak. I would like someone to verify my solution to this problem, on finding the minimizer/maximizer of a function of two variables, or any tips/hint to proceed ahead.

For each value of the scalar $\beta$, find the set of all stationary points of the following two variables $x$ and $y$
$$f(x,y) = x^2 + y^2 + \beta xy + x + 2y$$
Which of those stationary points are local minima? Which are global minima and why? Does this function have a global maximum for some value of $\beta$.

Solution.
We have:
\begin{align*}
f(x,y) &amp;= x^2 + y^2 + \beta xy + x +2y\\
f_x(x,y) &amp;= 2x+\beta y + 1\\
f_y(x,y) &amp;= \beta x + 2y + 2\\
f_{xx}(x,y) &amp;= 2\\
f_{xy}(x,y) &amp;= \beta \\
f_{yy}(x,y) &amp;= 2
\end{align*}
By the first order necessary condition(FONC) for optimality, we know that if $\nabla f(\mathbf{x})=0$, then $\mathbf{x}$ is a critical point.
Thus,
\begin{align*}
f_x(x,y) &amp;= 2x+\beta y + 1 = 0\\
f_y(x,y) &amp;= \beta x + 2y + 2 = 0
\end{align*}
Solving for $x$ and $y$, we find that:
\begin{align*}
x = \frac{\begin{array}{|cc|}
-1 &amp; \beta \\
-2 &amp; 2
\end{array}}{\begin{array}{|cc|}
2 &amp; \beta \\
\beta &amp; 2
\end{array}}=\frac{-2+2\beta}{4-\beta^2}=\frac{2\beta-2}{4 -\beta^2}
\end{align*}
\begin{align*}
y = \frac{\begin{array}{|cc|}
2 &amp; -1 \\
\beta &amp; -2
\end{array}}{\begin{array}{|cc|}
2 &amp; \beta \\
\beta &amp; 2
\end{array}}=\frac{-4+\beta}{4-\beta^2}=\frac{\beta -4}{4 - \beta^2}
\end{align*}
The second order necessary and sufficient conditions for optimality are based on the sign of the quadratic form $Q(\mathbf{h})=\mathbf{h}^T \cdot Hf(\mathbf{a}) \cdot \mathbf{h}$.
The Hessian of $f$ is given by,
$$Hf(\mathbf{x})=\begin{array}{|c c|}
2 &amp; \beta \\
\beta &amp; 2
\end{array}$$
Thus, $d_1 = 2 &gt; 0$ and $d_2 = 4 - \beta^2$. Thus, $f$ has a local minimizer if and only if $4 - \beta^2 &gt; 0$. $g(\beta) = 4 - \beta^2$ is a downward facing parabola. So, the values of this expression positive, if and only if $-2 &lt; \beta &lt; 2$. The function $f$ has no global maximum.
Question. How do I find the actual global minima?
If the function is convex, then all local minimum are global minimum.
If $4-\beta^2 &gt;0$, then the function is convex and hence the local minimum is indeed the global minimum.
If $\beta = \pm 2$, then we have $f(x,y)=(x\pm y)^2+x+2y$, $f(x, \mp x)=x\mp2x$ of which we can make it arbitrary large or small.
If $4-\beta^2 &lt; 0$, then it is indefinite, the stationary point is a saddle point.
It can't be negative definite, it doesn't have global maximum.