To answer the given question in a detailed and step-by-step manner:
### (a) List the basic and nonbasic variables
Basic variables are those associated with columns that have exactly one entry of 1 and the rest are 0s in the tableau. These columns correspond to identity matrix columns in a given row.
Nonbasic variables are the remaining variables not used for the basic variables.
Looking at the tableau:
[tex]\[
\left[\begin{array}{cccccccc}
x_1 & x_2 & x_3 & s_1 & s_2 & s_3 & z & \\
0 & 8 & 0 & \frac{1}{2} & 1 & \frac{4}{3} & 0 & 8 \\
1 & 5 & 0 & 5 & 0 & -7 & 0 & 12 \\
0 & 5 & 1 & \frac{7}{2} & 0 & -\frac{2}{3} & 0 & 26 \\
\hline
0 & 4 & 0 & \frac{1}{2} & 0 & 5 & 1 & 17
\end{array}\right]
\][/tex]
From the tableau, the columns for [tex]\( x_1 \)[/tex], [tex]\( x_3 \)[/tex], [tex]\( s_3 \)[/tex], and [tex]\( z \)[/tex] fit the criteria of having exactly one entry of 1 and the rest being 0s.
Thus, the basic variables are:
[tex]\[
\boxed{x_1, x_3, s_3, z}
\][/tex]
The nonbasic variables, which are the remaining variables, are:
[tex]\[
\boxed{x_2, s_1, s_2}
\][/tex]
### (b) Find the basic feasible solution
To determine the basic feasible solution, set the nonbasic variables to 0. This means:
[tex]\[
x_2 = 0, \quad s_1 = 0, \quad s_2 = 0
\][/tex]
Now, extract the values of the basic variables directly from the tableau.
- The value corresponding to [tex]\( s_3 \)[/tex] can be found where the column for [tex]\( s_3 \)[/tex] intersects the rightmost column of the tableau, which is 8.
- Similarly, [tex]\( x_1 \)[/tex] intersects at 12.
- For [tex]\( x_3 \)[/tex], the intersection is at 26.
- For [tex]\( z \)[/tex], the intersection is at 17.
Thus, the basic feasible solution is:
[tex]\[
\boxed{\{x_1: 12, x_3: 26, s_3: 8, z: 17\}}
\][/tex]
### (c) Determine whether this is a maximum solution
To decide whether this is a maximum solution, examine the coefficients in the bottom row (not including the rightmost value of the row):
[tex]\[0 \quad 4 \quad 0 \quad \frac{1}{2} \quad 0 \quad 5 \quad 1 \quad 17\][/tex]
For this to represent a maximum solution in the context of a maximization problem, all the entries in the bottom row (excluding the rightmost value) must be greater than or equal to 0.
The bottom row values are:
[tex]\[4, \quad 0.5, \quad 0, \quad 5, \quad 1\][/tex]
Since each of these values is greater than or equal to 0, this confirms that the current solution is indeed a maximum solution.
Therefore, the answer here is:
[tex]\[
\boxed{\text{True}}
\][/tex]
In summary:
- (a) The basic variables are [tex]\( x_1, x_3, s_3, z \)[/tex].
- (b) The basic feasible solution is [tex]\( \{x_1: 12, x_3: 26, s_3: 8, z: 17\} \)[/tex].
- (c) This is a maximum solution.
