How to derive the dual problem of Knapsack problem Knapsack problem is 
$$ \text{max} \, v^Tx$$
$$ \text{s.t.} \, w^Tx \le W, \, \, 0\le x_i \le 1 \, \, (i=1,...,n)$$
This is equivalent to 
$$   \text{min} \, -v^Tx$$
$$ \text{s.t.} \, {\left[\begin{array}{r}
   w^T \\
    I
  \end{array}\right]} x \le  {\left[\begin{array}{r}
   W\\
    e
  \end{array}\right]}  , \, \, x\ge 0 , \, \, e=(1,...,1)^T$$
Convert this into standard canonical form.
$$   \text{min} \, -v^Tx$$
$$ \text{s.t.} \, {\left[\begin{array}{r}
   w^T \\
    I
  \end{array}\right]}+y= {\left[\begin{array}{r}
   W\\
    e
  \end{array}\right]} 
$$
  $$  x, y \ge 0$$
$$ =  \text{min} \, -{\left[\begin{array}{r}
   v \\
    0 
  \end{array}\right]} ^T {\left[\begin{array}{r}
   x \\
    y
  \end{array}\right]} $$
$$ \text{s.t.} \, \, {\left(\begin{array}{rr}
   {\left[\begin{array}{r}
   w^T\\
    I
  \end{array}\right]} & I
  \end{array}\right)} {\left[\begin{array}{r}
   x \\
    y
  \end{array}\right]}=  {\left[\begin{array}{r}
   W\\
    e
  \end{array}\right]} $$
$$  x, y \ge 0$$
How can I find the dual of this?
I only know to solve up to here (I hope that upto here it is correct!). 
Answer should be
$$ \text{min} \, Wt+\sum_{i=1}^{n} S_i$$
$$ \text{s.t.} w_{i}t+S_i \ge v_i , \, (i=1,...,n)$$
$$t,s_1,...,s_n \ge 0 $$
I want to do this by using the following definition. The Standard form of the linear program is
 $$\text{Min} \,  C^{T} x$$
 $$ s.t.    Ax=b $$
 $$x\ge 0$$
Then the Dual is
 $$\text{Max}\, b^Ty$$
 $$s.t.   C-A^{T}y \ge 0$$
 A: *

*If you know that the dual of the dual is the primal problem itself, you don't need to rewrite the max problem into the min one.  Therefore, what's below the formulation of the LPP doesn't help to solve the problem.  Anyways, good try.

*It's possible that $\ge$ and $\le$ appear in the canonical form.


I'd prefer writing the primal problem (in the canonical form) in this way because it's more concise.
\begin{equation}
\begin{aligned}
\max z = v^T x & \\
\text{s.t. } w^T x &\le W \\
Ix &\le \mathbf{e} \\
x &\ge 0
\end{aligned}
\tag{*}
\label{primal}
\end{equation}
Here $\mathbf{e} = (1,1,\dots,1)^T \in \Bbb R^n$ and $W$ is a scalar.  If you don't see why the answer follows from \eqref{primal}, you may write it clearer like this.
\begin{equation}
\begin{aligned}
\max z = v^T x & \\
\text{s.t. } \begin{bmatrix} w^T \\ I \end{bmatrix} x &\le \begin{bmatrix} W \\ \mathbf{e} \end{bmatrix} \\
x &\ge 0
\end{aligned}
\tag{#}
\label{clear}
\end{equation}


*

*The dual variable $t \in \Bbb R$ represents the first constraint (row) in \eqref{clear}.

*The dual variable $s = (S_1,S_2,\dots,S_n)^T \in \Bbb R^n$ represents the rest  of the constraints in \eqref{clear}.


Therefore, the dual problem should be the last LPP stated in the OP.
\begin{align}
\min Z = \begin{bmatrix} W \\ \mathbf{e} \end{bmatrix}^T \begin{bmatrix} t \\ s \end{bmatrix} & \\
\text{s.t. } \begin{bmatrix} w & I \end{bmatrix} \begin{bmatrix} t \\ s \end{bmatrix} &\ge v \\
t,s &\ge 0
\end{align}
Simplify this to
\begin{align}
\min Z = W t + \mathbf{e}^T s & \\
\text{s.t. } w t + s &\ge v \\
t,s &\ge 0.
\end{align}
Write it in terms of scalar decision variables.
\begin{align}
\min Z = W t+\sum_{i=1}^{n} S_i & \\
\text{s.t. } w_{i}t+S_i &\ge v_i \, \forall i \in \{1,\dots,n\} \\
t,s_1,\dots,s_n &\ge 0 \, \forall i \in \{1,\dots,n\}
\end{align}

Finding the dual from the standard form
Note that $W \in \Bbb R, x,e \in \Bbb R^n, y \in \Bbb R^{n+1}$.  The original problem in standard form:
\begin{align}
\min z = -\begin{bmatrix} v \\ 0 \end{bmatrix}^T \begin{bmatrix} x \\ y \end{bmatrix} & \\
\text{s.t. } \left(\begin{array}{rr}
   {\left[\begin{array}{r}
   w^T\\
   I_n
  \end{array}\right]} & I_{n+1}
  \end{array}\right) \begin{bmatrix} x \\ y \end{bmatrix} &\ge \begin{bmatrix} W \\ e \end{bmatrix} \\
x,y &\ge 0
\end{align}
The dual is therefore
\begin{align}
\max Z = \begin{bmatrix} W \\ e \end{bmatrix}^T \begin{bmatrix} t \\ s \end{bmatrix} & \\
\text{s.t. } -\begin{bmatrix} v \\ 0 \end{bmatrix} -
\begin{bmatrix}
\begin{bmatrix} w & I_n \end{bmatrix} \\ I_{n+1}
\end{bmatrix}
\begin{bmatrix} t \\ s \end{bmatrix} &\ge 0
\end{align}
Write it using one more $\ge$ sign.
\begin{align}
\max Z = \begin{bmatrix} W \\ e \end{bmatrix}^T \begin{bmatrix} t \\ s \end{bmatrix} & \\
\text{s.t. } -v - \begin{bmatrix} w & I_n \end{bmatrix} \begin{bmatrix} t \\ s \end{bmatrix} &\ge 0 \\
-I_{n+1} \begin{bmatrix} t \\ s \end{bmatrix} &\ge 0
\end{align}
The last row means that $t,s \le 0$. Replace them by positive variables.
\begin{align}
\min Z = Wt + e^T s & \\
\text{s.t. } wt + s &\ge v \\
t,s &\ge 0
\end{align}
