1. **State the problem:** We want to maximize the objective function $$P = 30x + 12y$$ subject to the constraints $$3x + y \leq 18$$ and $$x, y \geq 0$$. 2. **Understand the constraints:** The inequalities define a feasible region in the first quadrant bounded by the line $$3x + y = 18$$. 3. **Find the intercepts of the constraint line:** - When $$x=0$$, $$y=18$$. - When $$y=0$$, $$x=6$$. 4. **Identify the corner points of the feasible region:** - $$A = (0,0)$$ - $$B = (6,0)$$ - $$C = (0,18)$$ 5. **Evaluate the objective function at each corner point:** - $$P(A) = 30(0) + 12(0) = 0$$ - $$P(B) = 30(6) + 12(0) = 180$$ - $$P(C) = 30(0) + 12(18) = 216$$ 6. **Conclusion:** The maximum value of $$P$$ is $$216$$ at the point $$C = (0,18)$$. Hence, the solution is $$x=0$$, $$y=18$$ with maximum $$P=216$$.