1. **State the problem:** We want to maximize the objective function $$P = 20x + 70y$$ subject to the constraints: $$x + 4y \leq 12$$ $$x \geq 0, \quad y \geq 0$$ 2. **Identify the feasible region:** The constraints define a region in the first quadrant bounded by the line $$x + 4y = 12$$ and the axes. 3. **Find the corner points of the feasible region:** - When $$x=0$$, $$4y=12 \Rightarrow y=3$$ - When $$y=0$$, $$x=12$$ - The third corner is at the origin $$(0,0)$$ 4. **Evaluate the objective function at each corner:** - At $$(0,0)$$: $$P=20(0)+70(0)=0$$ - At $$(12,0)$$: $$P=20(12)+70(0)=240$$ - At $$(0,3)$$: $$P=20(0)+70(3)=210$$ 5. **Determine the maximum value:** The maximum value of $$P$$ is $$240$$ at $$(12,0)$$. **Final answer:** $$\boxed{240}$$