Maximize f = 7x + 8y subject to x + 2y ≤ 10 x + y ≤ 7 . (x, y)

To solve this linear programming problem using the simplex method, we will follow these steps:

Step 1: Standard form

First, write the system in standard form by converting the inequalities to equalities with slack variables. The system is:

$\text{Maximize } f = 7x + 8y$

Subject to: $x + 2y \leq 10$ $x + y \leq 7$

To convert these inequalities into equalities, we introduce slack variables $s_1$ and $s_2$ for each inequality:

$x + 2y + s_1 = 10 \quad (1)$ $x + y + s_2 = 7 \quad (2)$

The objective function becomes:

$f = 7x + 8y$

Step 2: Set up the initial simplex tableau

We now set up the initial simplex tableau with the objective function and constraints.

\text{Basis} & x & y & s_1 & s_2 & \text{RHS} \\ \hline s_1 & 1 & 2 & 1 & 0 & 10 \\ s_2 & 1 & 1 & 0 & 1 & 7 \\ \hline f & -7 & -8 & 0 & 0 & 0 \\ \end{array}$$ ### Step 3: Choose entering and leaving variables - **Entering variable:** The most negative coefficient in the objective row is $$-8$$, so $$y$$ will enter the basis. - **Leaving variable:** Perform the ratio test to determine which variable will leave the basis. Divide the RHS by the corresponding values in the $$y$$ column: $$\frac{10}{2} = 5, \quad \frac{7}{1} = 7$$ Since 5 is the smallest, $$s_1$$ will leave the basis. ### Step 4: Pivot We now perform a pivot operation to make $$y$$ the new basic variable. We divide the first row by 2 to make the pivot element 1. New tableau after pivoting: $$\begin{array}{c|cccc|c} \text{Basis} & x & y & s_1 & s_2 & \text{RHS} \\ \hline y & \frac{1}{2} & 1 & \frac{1}{2} & 0 & 5 \\ s_2 & \frac{1}{2} & 0 & -\frac{1}{2} & 1 & 2 \\ \hline f & -3 & 0 & 4 & 0 & 40 \\ \end{array}$$ ### Step 5: Choose entering and leaving variables again - **Entering variable:** The most negative coefficient in the objective row is $$-3$$, so $$x$$ will enter the basis. - **Leaving variable:** Perform the ratio test again: $$\frac{5}{\frac{1}{2}} = 10, \quad \frac{2}{\frac{1}{2}} = 4$$ Since 4 is the smallest, $$s_2$$ will leave the basis. ### Step 6: Pivot again We perform another pivot operation to make $$x$$ the new basic variable. Divide the second row by $$\frac{1}{2}$$ to make the pivot element 1. New tableau after pivoting: $$\begin{array}{c|cccc|c} \text{Basis} & x & y & s_1 & s_2 & \text{RHS} \\ \hline y & 0 & 1 & 1 & -1 & 1 \\ x & 1 & 0 & -1 & 2 & 4 \\ \hline f & 0 & 0 & 1 & 6 & 52 \\ \end{array}$$ ### Step 7: Optimal solution There are no more negative coefficients in the objective row, so the current solution is optimal. The optimal values are: $$x = 4, \quad y = 1$$ The maximum value of the objective function is: $$f = 52$$ ### Final Answer: $$(x, y) = (4, 1)$$ $$f = 52$$ Would you like further clarification on any of the steps or more details on the simplex method? --- ### Relative questions: 1. How do you interpret the slack variables in a simplex tableau? 2. Can you explain the significance of the pivot operation in the simplex method? 3. Why is the ratio test important when choosing the leaving variable? 4. What happens if there is a tie in the ratio test? 5. How would you solve a minimization problem using the simplex method? ### Tip: When performing the ratio test, always discard non-positive values in the denominator to avoid infeasible results.