1. **State the problem:** Maximize the objective function $$P = 8x + 6y$$ subject to the constraints: $$x + 4y \geq 8$$ $$3x + 6y \geq 12$$ $$2x + y \leq 6$$ $$x \geq 0$$ $$y \geq 0$$ 2. **Identify the feasible region:** The feasible region is the set of points $(x,y)$ that satisfy all constraints simultaneously. 3. **Find the intersection points (vertices) of the feasible region:** - From $$x + 4y = 8$$, when $x=0$, $y=2$; when $y=0$, $x=8$. - From $$3x + 6y = 12$$, when $x=0$, $y=2$; when $y=0$, $x=4$. - From $$2x + y = 6$$, when $x=0$, $y=6$; when $y=0$, $x=3$. 4. **Calculate vertices by solving pairs of equations:** - Intersection of $$x + 4y = 8$$ and $$2x + y = 6$$: Multiply first equation by 2: $$2x + 8y = 16$$ Subtract second equation: $$(2x + 8y) - (2x + y) = 16 - 6$$ $$7y = 10$$ $$y = \frac{10}{7} \approx 1.43$$ Substitute back: $$x + 4 \times \frac{10}{7} = 8$$ $$x = 8 - \frac{40}{7} = \frac{56}{7} - \frac{40}{7} = \frac{16}{7} \approx 2.29$$ Vertex: $\left(\frac{16}{7}, \frac{10}{7}\right)$ - Intersection of $$3x + 6y = 12$$ and $$2x + y = 6$$: From second equation: $$y = 6 - 2x$$ Substitute into first: $$3x + 6(6 - 2x) = 12$$ $$3x + 36 - 12x = 12$$ $$-9x = -24$$ $$x = \frac{24}{9} = \frac{8}{3} \approx 2.67$$ Then: $$y = 6 - 2 \times \frac{8}{3} = 6 - \frac{16}{3} = \frac{18}{3} - \frac{16}{3} = \frac{2}{3} \approx 0.67$$ Vertex: $\left(\frac{8}{3}, \frac{2}{3}\right)$ - Intersection of $$x + 4y = 8$$ and $$3x + 6y = 12$$: Multiply first equation by 3: $$3x + 12y = 24$$ Subtract second equation: $$(3x + 12y) - (3x + 6y) = 24 - 12$$ $$6y = 12$$ $$y = 2$$ Substitute back: $$x + 4 \times 2 = 8$$ $$x = 8 - 8 = 0$$ Vertex: $(0, 2)$ 5. **Evaluate the objective function at each vertex:** - At $(0, 2)$: $$P = 8(0) + 6(2) = 12$$ - At $\left(\frac{16}{7}, \frac{10}{7}\right)$: $$P = 8 \times \frac{16}{7} + 6 \times \frac{10}{7} = \frac{128}{7} + \frac{60}{7} = \frac{188}{7} \approx 26.86$$ - At $\left(\frac{8}{3}, \frac{2}{3}\right)$: $$P = 8 \times \frac{8}{3} + 6 \times \frac{2}{3} = \frac{64}{3} + \frac{12}{3} = \frac{76}{3} \approx 25.33$$ - At $(0,6)$ (from $2x + y \leq 6$ when $x=0$): $$P = 8(0) + 6(6) = 36$$ 6. **Check feasibility of $(0,6)$:** - $x + 4y = 0 + 24 = 24 \geq 8$ ✓ - $3x + 6y = 0 + 36 = 36 \geq 12$ ✓ - $2x + y = 0 + 6 = 6 \leq 6$ ✓ - $x \geq 0$, $y \geq 0$ ✓ 7. **Conclusion:** The maximum value of $P$ is 36 at the vertex $(0,6)$. **Final answer:** $$\boxed{\max P = 36 \text{ at } (0,6)}$$