The initial tableau of a linear programming problem is given. Use the simplex method to solve the problem.
[tex]
\left[\begin{array}{rrrrrrr}x_1 & x_2 & s_1 & s_2 & s_3 & z & \\ 1 & 2 & 1 & 0 & 0 & 0 & 14 \\ 4 & 1 & 0 & 1 & 0 & 0 & 32 \\ 1 & 1 & 0 & 0 & 1 & 0 & 4 \\ \hline-4 & -3 & 0 & 0 & 0 & 1 & 0\end{array}\right]
[/tex]
The maximum is $\square$ when [tex]$x _1=$[/tex] $\square$ , [tex]$x_2=$[/tex] $\square$ , [tex]$s _1=$[/tex] $\square$ , [tex]$s _2=$[/tex] $\square$ , and [tex]$s _3=$[/tex] $\square$.
(Type integers or simplified fractions.)
Answer (2)
Identify the pivot column by finding the most negative entry in the last row.
Identify the pivot row by dividing the right-hand side values by the corresponding entries in the pivot column and choosing the smallest non-negative ratio.
Perform row operations to make the pivot element 1 and all other entries in the pivot column 0.
Extract the solution from the final tableau: z = 16 , x 1 = 4 , x 2 = 0 , s 1 = 10 , s 2 = 16 , s 3 = 0 .
Explanation
Understanding the Problem We are given the initial tableau of a linear programming problem and asked to use the simplex method to find the maximum value of z and the corresponding values of x 1 , x 2 , s 1 , s 2 , and s 3 .
Initial Tableau The initial tableau is: \left[\begin{array}{rrrrrrr}x_1 & x_2 & s_1 & s_2 & s_3 & z & \ 1 & 2 & 1 & 0 & 0 & 0 & 14 \ 4 & 1 & 0 & 1 & 0 & 0 & 32 \ 1 & 1 & 0 & 0 & 1 & 0 & 4 \ \hline-4 & -3 & 0 & 0 & 0 & 1 & 0\end{array}\right]
Finding the Pivot Column First, we identify the pivot column. This is the column with the most negative entry in the last row. In this case, it is the x 1 column with -4.
Finding the Pivot Row Next, we identify the pivot row. We divide the right-hand side values by the corresponding entries in the pivot column and choose the row with the smallest non-negative ratio. The ratios are: 14/1 = 14, 32/4 = 8, and 4/1 = 4. The pivot row is the third row, corresponding to s 3 .
Row Operations - First Iteration The pivot element is the element at the intersection of the pivot column and pivot row, which is 1. We perform row operations to make the pivot element 1 (it already is) and all other entries in the pivot column 0.
R1 = R1 - R3 R2 = R2 - 4 R3 R4 = R4 + 4 R3
New Tableau - First Iteration The new tableau is: \left[\begin{array}{rrrrrrr}x_1 & x_2 & s_1 & s_2 & s_3 & z & \ 0 & 1 & 1 & 0 & -1 & 0 & 10 \ 0 & -3 & 0 & 1 & -4 & 0 & 16 \ 1 & 1 & 0 & 0 & 1 & 0 & 4 \ \hline0 & 1 & 0 & 0 & 4 & 1 & 16\end{array}\right]
Checking for Optimality Now, we repeat the process. We look for the most negative entry in the last row. There are no negative entries, so we proceed to extract the solution.
Extracting the Solution From the final tableau, we can read the solution: z = 16 x 1 = 4 x 2 = 0 s 1 = 10 s 2 = 16 s 3 = 0
Final Answer Therefore, the maximum is 16 when x 1 = 4 , x 2 = 0 , s 1 = 10 , s 2 = 16 , and s 3 = 0 .
Examples
Linear programming is used in resource allocation to maximize profit or minimize cost. For example, a company might use linear programming to determine how much of each product to produce in order to maximize its profit, given constraints on the availability of resources such as labor and materials. The simplex method helps to solve these optimization problems efficiently.
Using the simplex method, we found the optimal solution for the linear programming problem. The maximum value of z is 16, occurring when x 1 = 4 , x 2 = 0 , s 1 = 10 , s 2 = 16 , and s 3 = 0 .
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