suppose we have
$$ min z = 6 x_1 + 20 x+3 $$
such that
$$ x_2 + 4 x_3 \leq 10 $$ $$ - x_2 +2 x_3 \leq 11 $$ $$ x_i \geq 0 $$
Find the dual.
ATTEMPT:
I wrote
$$ max = 10 y_1 + 11 y_2 $$
st
\begin{align*} 30 y_2 \geq 6 \\ y_1 - y_2 \geq 0 \\ 4y_1 + 2 y_2 \geq 20 \\ y_i \geq 0 \end{align*}
but the lecture notes say $y_i \leq 0$. I dont understand why this is so. probabtly is it a typo?
The lecturer is right. The constraints of your min problem are $\leq$, so the primal is equivalent to:
$$\begin{align} \min \quad & 6 x_1 + 20 x_3 \\ \text {s.t.} \quad & -x_2 - 4 x_3 \geq -10 \\ & -30 x_1 + x_2 -2 x_3 \geq -11 \\ & x_i \geq 0 \end{align}$$
whose dual is: $$\begin{align} \max \quad & -10 y_1 -11 y_2 \\ \text {s.t.} \quad & 30 y_2 \leq 6 \\ & -y_1 + y_2 \leq 0 \\ & - 4y_1 -2 y_2 \leq 20 \\ & y_i \geq 0 \end{align}$$ Now replace $y$ with $-y$ to obtain the solution of the teacher.