Compute the optimal point $x^*$ and $f^* = f(x^*)$?

Question

$$f(x) = max_i\{a_i^T x + b \}$$

with

$a_1 =\begin{pmatrix} 1\\ 0 \end{pmatrix}$, $a_2=\begin{pmatrix} -\frac{1}{\sqrt{5}}\\ \frac{2}{\sqrt{5}} \end{pmatrix}$, $a_3 = \begin{pmatrix} -\frac{1}{\sqrt{5}}\\ -\frac{2}{\sqrt{5}} \end{pmatrix}$, $b=-3$

I would use the Gaussian elimination and solve a system of 3 equations.

$$ \begin{matrix} a_1^T \cdot [x_1, x_2]^T & = & -3 \\ a_2^T \cdot [x_1, x_2]^T & = & -3 \\ a_3^T \cdot [x_1, x_2]^T & = & -3 \\ \end{matrix}$$

... and got as result: $x^*= [-3, \frac{-3\sqrt(5)-3}{2}]^T$

I am not sure, shouldn't be the optimal point at $(0,0)$?

Answer 1

Hint

$$ f(x) = \max_i\{a_i^{\top}x+b\} $$

$$ f(x) = \min_i\{a_i^{\top}x+b\} $$