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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?

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    $\begingroup$ Try setting some values at the parameter $betta$ and creating a 3D plot. It will help you build an intuition. $\endgroup$ May 9 '21 at 6:53
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    $\begingroup$ This question is appropriate for math.SE I belive $\endgroup$
    – 10xAI
    May 9 '21 at 16:26
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If the function is convex, then all local minimum are global minimum.

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.

It can't be negative definite, it doesn't have global maximum.

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