1. **Problem Statement:** Maximize the objective function $$30x + 50y$$ subject to the constraints: $$3x + 4y \geq 12$$ $$4x + 2y \leq 20$$ $$x \leq 4$$ $$y \geq 6$$ $$x \geq 0$$ $$y \geq 0$$ 2. **Method:** This is a linear programming problem. The maximum value of the objective function occurs at a vertex (corner point) of the feasible region defined by the constraints. 3. **Find the vertices of the feasible region:** - From $$x \geq 0$$ and $$y \geq 0$$, we are in the first quadrant. - The constraints form boundary lines: - Line 1: $$3x + 4y = 12$$ - Line 2: $$4x + 2y = 20$$ - Line 3: $$x = 4$$ - Line 4: $$y = 6$$ 4. **Calculate intersections of these lines within the feasible region:** - Intersection of Line 1 and Line 2: Solve system: $$3x + 4y = 12$$ $$4x + 2y = 20$$ Multiply second equation by 2: $$8x + 4y = 40$$ Subtract first equation: $$8x + 4y - (3x + 4y) = 40 - 12$$ $$5x = 28 \Rightarrow x = \frac{28}{5} = 5.6$$ Substitute back: $$3(5.6) + 4y = 12 \Rightarrow 16.8 + 4y = 12 \Rightarrow 4y = -4.8 \Rightarrow y = -1.2$$ Since $$y = -1.2 < 0$$, this point is outside the feasible region. - Intersection of Line 1 and Line 3 ($$x=4$$): $$3(4) + 4y = 12 \Rightarrow 12 + 4y = 12 \Rightarrow 4y = 0 \Rightarrow y = 0$$ Point: $$(4,0)$$ but $$y \geq 6$$ is required, so this is outside feasible region. - Intersection of Line 1 and Line 4 ($$y=6$$): $$3x + 4(6) = 12 \Rightarrow 3x + 24 = 12 \Rightarrow 3x = -12 \Rightarrow x = -4$$ $$x = -4 < 0$$, outside feasible region. - Intersection of Line 2 and Line 3 ($$x=4$$): $$4(4) + 2y = 20 \Rightarrow 16 + 2y = 20 \Rightarrow 2y = 4 \Rightarrow y = 2$$ $$y = 2 < 6$$, outside feasible region. - Intersection of Line 2 and Line 4 ($$y=6$$): $$4x + 2(6) = 20 \Rightarrow 4x + 12 = 20 \Rightarrow 4x = 8 \Rightarrow x = 2$$ Point: $$(2,6)$$ satisfies all constraints. - Intersection of Line 3 and Line 4: $$(4,6)$$ 5. **Check feasibility of points:** - $$(2,6)$$ satisfies: $$3(2) + 4(6) = 6 + 24 = 30 \geq 12$$ $$4(2) + 2(6) = 8 + 12 = 20 \leq 20$$ $$x=2 \leq 4$$ $$y=6 \geq 6$$ $$x,y \geq 0$$ - $$(4,6)$$ satisfies: $$3(4) + 4(6) = 12 + 24 = 36 \geq 12$$ $$4(4) + 2(6) = 16 + 12 = 28 \not\leq 20$$ (violates constraint) - $$(4,0)$$ violates $$y \geq 6$$ - $$(0,6)$$ check: $$3(0) + 4(6) = 24 \geq 12$$ $$4(0) + 2(6) = 12 \leq 20$$ $$x=0 \leq 4$$ $$y=6 \geq 6$$ - $$(0,0)$$ violates $$y \geq 6$$ 6. **Vertices in feasible region:** $$(0,6), (2,6)$$ 7. **Evaluate objective function at vertices:** - At $$(0,6)$$: $$30(0) + 50(6) = 300$$ - At $$(2,6)$$: $$30(2) + 50(6) = 60 + 300 = 360$$ 8. **Maximum value:** $$360$$ at $$(2,6)$$ **Final answer:** The maximum value of $$30x + 50y$$ subject to the constraints is $$\boxed{360}$$ at $$(x,y) = (2,6)$$.