Consider the problem $$ \begin{array}{ll} \underset {x_1, x_2, x_3} {\text{maximize}} & c_1 x_1 + c_2 x_2 + c_3 x_3 \\ \text{subject to} & x_1 + 2 x_2 + 3 x_3 \leq 6 \\ & x_1 \geq 0 \\ & x_2 \geq 0 \\ & 1 \geq x_3 \geq 0 \end{array} $$ Convert this problem to an equivalent problem in standard form. Write your final answer in the form $Ax = b$, $x \geq 0$.
Would the result be A=[1 2 3 1] x = [x1 x2 x3 x4 u] b = 6? Or do I have to do something about the 1 ≥ x3 ≥ 0 condition?
So the result would be
$$ A = \begin{bmatrix}1&2&3&1&0\\0&0&1&0&1\end{bmatrix}, \qquad x = \begin{bmatrix}x1\\x2\\x3\\x4\\u\\v\end{bmatrix}, \qquad \qquad b = \begin{bmatrix}6\\1\end{bmatrix} $$
It is almost correct. The dimension of $\textbf A$ and $\textbf x$ do not match. The correct matrix notation is
$$\texttt{Max} \ \ \begin{pmatrix} c_1 & c_2&c_3 \end{pmatrix}\cdot \begin{pmatrix} x_1 \\ x_2\\x_3 \end{pmatrix}$$
$$\underbrace{\begin{pmatrix}1&2&3&1&0\\0&0&1&0&1\end{pmatrix}}_{\color{red}2\times 5 \\ \\ A}\cdot \underbrace{\begin{pmatrix}x_1\\x_2\\x_3\\u\\v\end{pmatrix}}_{5\times \color{blue}1 \\ \textbf { x}}=\underbrace{\begin{pmatrix}6\\1\end{pmatrix}}_{\color{red}2\times \color{blue}1 \\ \textbf b}$$
$$\texttt{Non-negativity condition: }\textbf x\geq 0$$