1. **State the problem:** Maximize the objective function $$z = 3x + 7y$$ subject to the constraints: $$4x + 6y \leq 24$$ $$9x + y \leq 24$$ $$x \geq 0, y \geq 0$$ 2. **Identify the feasible region:** The feasible region is the set of points $(x,y)$ that satisfy all constraints simultaneously. It is bounded by the lines: - $$4x + 6y = 24$$ - $$9x + y = 24$$ - The coordinate axes $$x=0$$ and $$y=0$$ 3. **Find the vertices of the feasible region:** The maximum of a linear function over a polygonal region occurs at a vertex. - Vertex A: Intersection of $$x=0$$ and $$4x + 6y = 24$$ Substitute $$x=0$$ into $$4(0) + 6y = 24$$ gives $$6y=24 \Rightarrow y=4$$ So, A = $$(0,4)$$ - Vertex B: Intersection of $$9x + y = 24$$ and $$4x + 6y = 24$$ Solve the system: $$9x + y = 24$$ $$4x + 6y = 24$$ From the first, $$y = 24 - 9x$$ Substitute into second: $$4x + 6(24 - 9x) = 24$$ $$4x + 144 - 54x = 24$$ $$-50x = 24 - 144 = -120$$ $$x = \frac{-120}{-50} = \frac{12}{5} = 2.4$$ Then $$y = 24 - 9(2.4) = 24 - 21.6 = 2.4$$ So, B = $$(2.4, 2.4)$$ - Vertex C: Intersection of $$y=0$$ and $$9x + y = 24$$ Substitute $$y=0$$ into $$9x + 0 = 24$$ gives $$x = \frac{24}{9} = \frac{8}{3} \approx 2.67$$ So, C = $$(2.67, 0)$$ - Vertex D: Intersection of $$y=0$$ and $$4x + 6y = 24$$ Substitute $$y=0$$ into $$4x + 0 = 24$$ gives $$x=6$$ So, D = $$(6,0)$$ 4. **Evaluate the objective function at each vertex:** - At A: $$z = 3(0) + 7(4) = 28$$ - At B: $$z = 3(2.4) + 7(2.4) = 7.2 + 16.8 = 24$$ - At C: $$z = 3(2.67) + 7(0) = 8.01$$ - At D: $$z = 3(6) + 7(0) = 18$$ 5. **Determine the maximum value:** The maximum value of $$z$$ is $$28$$ at vertex A $$(0,4)$$. **Final answer:** $$\boxed{\text{Maximum value } z = 28 \text{ at } (x,y) = (0,4)}$$