# Optimal points of $f(x,y)=x^2 + y^2 + \beta xy + x + 2y$

I am self-learning basic optimization theory and algorithms from "An Introduction to Optimization" 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) &= x^2 + y^2 + \beta xy + x +2y\\ f_x(x,y) &= 2x+\beta y + 1\\ f_y(x,y) &= \beta x + 2y + 2\\ f_{xx}(x,y) &= 2\\ f_{xy}(x,y) &= \beta \\ f_{yy}(x,y) &= 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) &= 2x+\beta y + 1 = 0\\ f_y(x,y) &= \beta x + 2y + 2 = 0 \end{align*}

Solving for $$x$$ and $$y$$, we find that:

\begin{align*} x = \frac{\begin{array}{|cc|} -1 & \beta \\ -2 & 2 \end{array}}{\begin{array}{|cc|} 2 & \beta \\ \beta & 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 & -1 \\ \beta & -2 \end{array}}{\begin{array}{|cc|} 2 & \beta \\ \beta & 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,

$$Hf(\mathbf{x})=\begin{array}{|c c|} 2 & \beta \\ \beta & 2 \end{array}$$

Thus, $$d_1 = 2 > 0$$ and $$d_2 = 4 - \beta^2$$. Thus, $$f$$ has a local minimizer if and only if $$4 - \beta^2 > 0$$. $$g(\beta) = 4 - \beta^2$$ is a downward facing parabola. So, the values of this expression positive, if and only if $$-2 < \beta < 2$$. The function $$f$$ has no global maximum.

Question. How do I find the actual global minima?

• Try setting some values at the parameter $betta$ and creating a 3D plot. It will help you build an intuition. May 9 '21 at 6:53
• This question is appropriate for math.SE I belive May 9 '21 at 16:26

If $$4-\beta^2 >0$$, 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 $$4-\beta^2 < 0$$, then it is indefinite, the stationary point is a saddle point.