1. **State the problem:** Maximize the objective function $$P = 30x + 12y$$ subject to the constraints $$3x + y \leq 18$$ and $$x, y \geq 0$$. 2. **Convert inequalities to equations by adding slack variable:** Introduce slack variable $$s \geq 0$$ to convert $$3x + y \leq 18$$ into $$3x + y + s = 18$$. 3. **Set up the initial simplex tableau:** $$\begin{array}{c|ccc|c} & x & y & s & \text{RHS} \\ \hline \text{Constraint} & 3 & 1 & 1 & 18 \\ \hline \text{Objective (P)} & -30 & -12 & 0 & 0 \end{array}$$ 4. **Identify entering variable:** The most negative coefficient in the objective row is $$-30$$ for $$x$$, so $$x$$ enters the basis. 5. **Identify leaving variable:** Calculate ratios $$\frac{18}{3} = 6$$ for slack variable $$s$$ row. Since only one constraint, $$s$$ leaves the basis. 6. **Pivot to make pivot element 1:** Divide the constraint row by 3: $$\begin{array}{c|ccc|c} & x & y & s & \text{RHS} \\ \hline \text{Constraint} & 1 & \frac{1}{3} & \frac{1}{3} & 6 \\ \hline \text{Objective (P)} & -30 & -12 & 0 & 0 \end{array}$$ 7. **Eliminate $$x$$ from objective row:** Add 30 times the constraint row to the objective row: $$\begin{aligned} &-30 + 30 \times 1 = 0 \\ &-12 + 30 \times \frac{1}{3} = -12 + 10 = -2 \\ &0 + 30 \times \frac{1}{3} = 10 \\ &0 + 30 \times 6 = 180 \end{aligned}$$ Updated tableau: $$\begin{array}{c|ccc|c} & x & y & s & \text{RHS} \\ \hline \text{Constraint} & 1 & \frac{1}{3} & \frac{1}{3} & 6 \\ \hline \text{Objective (P)} & 0 & -2 & 10 & 180 \end{array}$$ 8. **Repeat:** The most negative coefficient in objective row is $$-2$$ for $$y$$, so $$y$$ enters the basis. 9. **Identify leaving variable:** Calculate ratio for constraint row: $$\frac{6}{\frac{1}{3}} = 18$$. 10. **Pivot to make pivot element 1:** Multiply constraint row by 3: $$\begin{array}{c|ccc|c} & x & y & s & \text{RHS} \\ \hline \text{Constraint} & 3 & 1 & 1 & 18 \\ \hline \text{Objective (P)} & 0 & -2 & 10 & 180 \end{array}$$ 11. **Eliminate $$y$$ from objective row:** Add 2 times constraint row to objective row: $$\begin{aligned} &0 + 2 \times 3 = 6 \\ &-2 + 2 \times 1 = 0 \\ &10 + 2 \times 1 = 12 \\ &180 + 2 \times 18 = 216 \end{aligned}$$ Updated tableau: $$\begin{array}{c|ccc|c} & x & y & s & \text{RHS} \\ \hline \text{Constraint} & 3 & 1 & 1 & 18 \\ \hline \text{Objective (P)} & 6 & 0 & 12 & 216 \end{array}$$ 12. **Check for optimality:** No negative coefficients in objective row, so optimal solution reached. 13. **Read solution:** From constraint row, $$3x + y + s = 18$$ with $$s=0$$ (slack variable), and from objective row, $$y$$ is basic variable with zero coefficient in objective row. Since $$y$$ entered last, solve for $$y$$: $$y = 18 - 3x$$. From step 11, objective row coefficient for $$x$$ is positive, so $$x=0$$ to maximize $$P$$. Therefore, $$x=0$$, $$y=18$$, and maximum $$P=216$$. **Final answer:** $$x=0, y=18, P_{max}=216$$.