Automatic Control (AC)

Question 3: Optimization

August 2016 Problem 3

Solution

Let $t_1=x_1-2$, $t_2=x_2+1$
so that $g(t_1,t_2)=\dfrac{1}{t_1^2+t_2^2+3}|t_1=0,t_2=0$ would have some convex property
with $f(x_1,x_2)=\dfrac{1}{(x_1-2)^2+(x_2+1)^2+3}|x_1=2,x_1=-1$
$D^2g(x)=\dfrac{1}{(t_1^2+t_2^2+3)^3}\begin{bmatrix} 6(t_1)^2-2(t_2)^3-6 & 8t_1t_2 \\ 8t_1t_2 & 6(t_2)^2-2(t_1)^3-6 \end{bmatrix}=\dfrac{1}{27}\begin{bmatrix} -6 & 0 \\ 0 & -6 \end{bmatrix}$
It is easy to see that $\begin{bmatrix} -6 & 0 \\ 0 & -6 \end{bmatrix}$ is n.d .
Such that function at $[2 -1]^T$ is strictly locally concave.

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