Assumptions
- $A \in M_{m\times n}(\Bbb R)$ ($m\le n$) has rank $n$ and basis matrix $B$.
- $x_B$ denotes the basic solution.
- $c_B$ denotes the reduced objective function so that $c^T x=c_B^T x_B$. (So the order/arrangement of basic variables is very important.)
The initial simplex tableau for your problem(, which is already in canoncial form,) will be
\begin{array}{cc|c}
A & I & b \\ \hline
-c^T & 0 & 0
\end{array}
We multiply the $m$ constraints by $B^{-1}$. Note that $\mathbf{e}_1,\dots,\mathbf{e}_n$ can be found in $[B^{-1}A\;B^{-1}]$ since $B$ is chosen from columns of $[A\;I]$. We add a linear combination of the first $m$ row to the last row.
\begin{array}{cc|c}
B^{-1}A & B^{-1} & B^{-1}b \\ \hline
c_B^T B^{-1}A-c^T & c_B^T B^{-1} & c_B^T B^{-1}b \tag{*} \label{t1}
\end{array}
Why is $c_B^T$ chosen? We write $x_B$ as $(x_{B_1},\dots,x_{B_n})^T$. Choose an index $i$ in $\{1,\dots,n\}$, and suppose that $x_j = x_{B_i}$
Case 1: $1\le j \le n$. In the $j$-th column of the last row,
$$(c_B^T B^{-1}A-c^T)_j = c_B^T B^{-1} \mathbf{a}_j -c_j = c_B^T \mathbf{e}_i - c_j = c_{B_i} - c_j = 0.$$
Case 2: $n+1 \le j \le m+n$. Note that $c_{B_i}= 0$. In the $j$-th column of the last row,
$$(c_B^T B^{-1})_{j - n} = c_B^T \mathbf{e}_i = c_{B_i} = 0.$$
Therefore, the entries corresponding to the basic variables in the last row in tableau $\eqref{t1}$ will be zero. Then, $\eqref{t1}$ is a simplex tableau that you can compute from the given optimal solution.
Computational results for your example
Here's the GNU Octave code for finding the optimal tableau. The syntax should be the same in MATLAB except for the solver glpk. You may test the code at Octave Online.
c=[60 100 80]'; A=[144 192 240; 100 150 120; 1 1 1];
b=[120000 60000 500]';
[x_max,z_max]=glpk(c,A,b,[0 0 0]',[],"UUU","CCC",-1)
x=[x_max; b-A*x_max] % Include slack variables
A1=[A eye(3)]; % Part of initial tableau
c1=[c' zeros(1,3)]';
B=A1(:,find(x > 0)')
c_B=c1(find(x > 0));
disp('Initial tableau');
[A1 b; -c1' 0]
disp('Final tableau');
[B^-1*A B^-1 B^-1*b; c_B'*B^-1*A-c' c_B'*B^-1 c_B'*B^-1*b]
The computer output
x_max =
0.00000
400.00000
0.00000
z_max = 4.0000e+04
x =
0.0000e+00
4.0000e+02
0.0000e+00
4.3200e+04
-7.2760e-12
1.0000e+02
B =
192 1 0
150 0 0
1 0 1
Initial tableau
ans =
144 192 240 1 0 0 120000
100 150 120 0 1 0 60000
1 1 1 0 0 1 500
-60 -100 -80 -0 -0 -0 0
Final tableau
ans =
Columns 1 through 6:
6.6667e-01 1.0000e+00 8.0000e-01 0.0000e+00 6.6667e-03 -0.0000e+00
1.6000e+01 0.0000e+00 8.6400e+01 1.0000e+00 -1.2800e+00 0.0000e+00
3.3333e-01 1.1102e-16 2.0000e-01 0.0000e+00 -6.6667e-03 1.0000e+00
6.6667e+00 0.0000e+00 0.0000e+00 0.0000e+00 6.6667e-01 0.0000e+00
Column 7:
4.0000e+02
4.3200e+04
1.0000e+02
4.0000e+04
The optimal tableau:
\begin{equation*}
\begin{array}{ccccccc|l}
& x_1 & x_2 & x_3 & s_1 & s_2 & s_3 & \\ \hline
x_2 & \frac23 & 1 & \frac45 & 0 & \frac{1}{150} & 0 & 400 \\
s_1 & 16 & 0 & \frac{432}{5} & 1 & -\frac{32}{25} & 0 & 43200 \\
s_3 & \frac13 & 0 & \frac15 & 0 & -\frac{1}{150} & 1 & 100 \\ \hline
& \frac{20}{3} & 0 & 0 & 0 & \frac23 & 0 & 40000 \\
\end{array}
\end{equation*}
After you have minimized the LP, there are no more reduced costs, e.g. they are all zero. All that you have are the dual variables. You only do have nonzero reduced costs at a local minimum (=basis = edge of the feasible polyhedron).
Generally, the reduced cost is:
\begin{equation}
d_j = c_j - \mu^\top A_j
\end{equation}
Cost of increasing a nonbasic variable $x_j$. Nonbasic row of the coefficientmatrix $A$: $A_j$.
With dual variable $\mu$:
\begin{equation}
\mu = c_B^\top B^{-1}
\end{equation}
$B$ is the Basis of coefficient matrix $A$, sometimes $B$ is called $A_B$.
There is an example in Bertsimas - Introduction to Linear Optimization on p.84.
Best Answer
Hint
You can't find the quantity you're looking for without making some assumption about the original objective function. Presumably, the assumption you're supposed to make is that the original objective was a function of $\ x_1\ $ and $\ x_2\ $ only (which would be the case, for instance, if $\ s_1\ $ and $\ s_2\ $ were slack variables). If that's the case, then you can find the original objective function by performing row operations on the tableau to eliminate the entries in the third and fourth columns of the first row. The entries in the first and second columns of the first row will then be the coefficients of $\ x_1\ $ and $\ x_2\ $ in the original objective function. Since you know that the optimal solution is $\ x_1=4, x_2=9\ $, you can now obtain the value of the objective function by substituting those values in the expression you've just obtained for it.