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Problem 10

$\begin{array}{ll}\text { Maximize } & z=3 x_{1}+4 x_{2}+6 x_{3} \\ \text { subject to } & 5 x_{1}-x_{2}+x_{3} \leq 1,500 \\ & 2 x_{1}+2 x_{2}+x_{3} \leq 2,500 \\ & 4 x_{1}+2 x_{2}+x_{3} \leq 2,000 \\ & x_{1} \geq 0, x_{2} \geq 0, x_{3} \geq 0\end{array}$

Expert verified
The optimal solution for the given linear programming problem is $$x_1 = 0$$, $$x_2 = 0$$, $$x_3 = 1500$$, with a maximum value for $$z$$ of 9000.
See the step by step solution

Step 1: Identify the objective function and constraints

Objective function: Maximize $$z = 3x_1 + 4x_2 + 6x_3$$ Constraints: $$5x_1 - x_2 + x_3 \leq 1500$$ $$2x_1 + 2x_2 + x_3 \leq 2500$$ $$4x_1 + 2x_2 + x_3 \leq 2000$$ $$x_1 \geq 0, x_2 \geq 0, x_3 \geq 0$$

Step 2: Add slack variables to the constraints

Introduce slack variables, $$s_1, s_2, s_3$$ to convert the inequality constraints into equalities: $$5x_1 - x_2 + x_3 + s_1 = 1500$$ $$2x_1 + 2x_2 + x_3 + s_2 = 2500$$ $$4x_1 + 2x_2 + x_3 + s_3 = 2000$$ Slack variables, $$s_1, s_2, s_3 \geq 0$$

Step 3: Set up the initial simplex tableau

$$\begin{array}{ccccccc|c} c_j & 3 & 4 & 6 & 0 & 0 & 0 & \\ \hline B & x_1 & x_2 & x_3 & s_1 & s_2 & s_3 & b \\ \hline s_1 & 5 & -1 & 1 & 1 & 0 & 0 & 1500 \\ s_2 & 2 & 2 & 1 & 0 & 1 & 0 & 2500 \\ s_3 & 4 & 2 & 1 & 0 & 0 & 1 & 2000 \\ \hline z = & -3 & -4 & -6 & 0 & 0 & 0 & 0 \\ \end{array}$$

Step 4: Perform pivot operations

The most negative relative cost in row $$z$$ corresponds to $$x_3$$, so we pivot on this column. To find the pivot row, we use the minimum ratio test: $$\frac{1500}{1} = 1500 , \frac{2500}{1} = 2500, \frac{2000}{1} = 2000$$ The pivot element is in the first row and third column. Perform pivot operations to get the updated simplex tableau: $$\begin{array}{ccccccc|c} c_j & 3 & 4 & 6 & 0 & 0 & 0 & \\ \hline B & x_1 & x_2 & x_3 & s_1 & s_2 & s_3 & b \\ \hline x_3 & 5 & -1 & 1 & 1 & 0 & 0 & 1500\\ s_2 & -3 & 4 & 0 & -1 & 1 & 0 & 1000 \\ s_3 & -1 & 3 & 0 & -1 & 0 & 1 & 500 \\ \hline z = & 9 & 2 & 0 & 6 & 0 & 0 & 9000 \\ \end{array}$$

Step 5: Identify the optimal solution

All relative costs in row $$z$$ are non-negative, indicating we've found the optimal solution. The optimal solution for this LP problem is: $$x_1 = 0, x_2 = 0, x_3 = 1500, z = 9000$$

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